Class 12 RD Sharma Solutions - Chapter 20 Definite Integrals - Exercise 20.2 | Set 1

Evaluate the following definite integrals:

Question 1. \int_{2}^{4}\frac{x}{x^2+1}dx

Solution:

We have,

I = \int_{2}^{4}\frac{x}{x^2+1}dx

I = \frac{1}{2}\int_{2}^{4}\frac{2x}{x^2+1}dx

I = \left[\frac{1}{2}\log(1+x^2)\right]_{2}^{4}

I = \frac{1}{2}\log(1+4^2)-\frac{1}{2}\log(1+2^2)

I = \frac{1}{2}\log(1+16)-\frac{1}{2}\log(1+4)

I = \frac{1}{2}\log17-\frac{1}{2}\log5

I = \frac{1}{2}(\log17-log5)

I = \frac{1}{2}\log\frac{17}{5}

Therefore, the value of \int_{2}^{4}\frac{x}{x^2+1}dx is \frac{1}{2}\log\frac{17}{5}.

Question 2. \int_{1}^{2}\frac{1}{x(1+logx)^2}dx

Solution:

We have,

I = \int_{1}^{2}\frac{1}{x(1+logx)^2}dx

Let 1 + log x = t, so we have,

=> (1/x) dx = 2t dt

Now, the lower limit is, x = 1

=> t = 1 + log x

=> t = 1 + log 1

=> t = 1 + 0

=> t = 1

Also, the upper limit is, x = 2

=> t = 1 + log x

=> t = 1 + log 2

So, the equation becomes,

I = \int_{1}^{1+\log2}\frac{1}{t^2}dt

I = \left[\frac{-1}{t}\right]_1^{1+\log2}

I = \left[\frac{-1}{1+\log2}+1\right]

I = \frac{-1+1+\log2}{1+\log2}

I = \frac{\log2}{1+\log2}

I = \frac{\log2}{\log e+\log2}

I = \frac{\log2}{\log2e}

Therefore, the value of \int_{1}^{2}\frac{1}{x(1+logx)^2}dx is \frac{\log2}{\log2e}.

Question 3. \int_{1}^{2}\frac{3x}{9x^2-1}dx

Solution:

We have,

I = \int_{1}^{2}\frac{3x}{9x^2-1}dx

Let 9x² – 1 = t, so we have,

=> 18x dx = dt

=> 3x dx = dt/6

Now, the lower limit is, x = 1

=> t = 9x² – 1

=> t = 9 (1)² – 1

=> t = 9 – 1

=> t = 8

Also, the upper limit is, x = 2

=> t = 9x² – 1

=> t = 9 (2)² – 1

=> t = 36 – 1

=> t = 35

So, the equation becomes,

I = \int_{8}^{35}\frac{1}{6t}dt

I = \left[\frac{1}{6}\log t\right]_{8}^{35}

I = \frac{1}{6}(\log 35-\log 8)

I = \frac{1}{6}\log \frac{35}{8}

Therefore, the value of \int_{1}^{2}\frac{3x}{9x^2-1}dx is \frac{1}{6}\log \frac{35}{8}.

Question 4. \int_{0}^{\frac{\pi}{2}}\frac{1}{5cosx+3sinx}dx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}}\frac{1}{5cosx+3sinx}dx

On putting sin x = \frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}} and cos x = \frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}, we get

I = \int_{0}^{\frac{\pi}{2}}\frac{sec^2\frac{x}{2}}{5(1-tan^2\frac{x}{2})+6tan\frac{x}{2}}dx

I = \int_{0}^{\frac{\pi}{2}}\frac{sec^2\frac{x}{2}}{5-5tan^2\frac{x}{2}+6tan\frac{x}{2}}dx

Let tan x/2 = t. So, we have

=> \frac{1}{2}sec^2\frac{x}{2} = dt

Now, the lower limit is, x = 0

=> t = tan x/2

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = tan x/2

=> t = tan π/4

=> t = 1

So, the equation becomes,

I = \int_{0}^{1}\frac{2}{5-5t^2+6t}dt

I = \frac{2}{5}\int_{0}^{1}\frac{1}{1-t^2+\frac{6}{5}t}dt

I = \frac{2}{5}\int_{0}^{1}\frac{1}{\frac{34}{25}-(t-\frac{3}{5})^2}dt

I = \left[\frac{2}{5}.\frac{1}{2}\sqrt{\frac{25}{34}}\log\left(\frac{\sqrt{\frac{34}{25}}+t-\frac{3}{5}}{\sqrt{\frac{34}{25}}-t+\frac{3}{5}}\right)\right]^1_0

I = \left[\frac{2}{5}.\frac{1}{2}\frac{5}{\sqrt{34}}\log\left(\frac{\sqrt{\frac{34}{25}}+t-\frac{3}{5}}{\sqrt{\frac{34}{25}}-t+\frac{3}{5}}\right)\right]^1_0

I = \frac{1}{\sqrt{34}}\left[\log\left(\frac{\sqrt{34}+2}{\sqrt{34}-2}\right)-\log\left(\frac{\sqrt{34}-3}{\sqrt{34}+3}\right)\right]

I = \frac{1}{\sqrt{34}}\log\left(\frac{(\sqrt{34}+2)(\sqrt{34}+3)}{(\sqrt{34}-2)(\sqrt{34}-3)}\right)

I = \frac{1}{\sqrt{34}}\log\left(\frac{40+5\sqrt{34}}{40-5\sqrt{34}}\right)

I = \frac{1}{\sqrt{34}}\log\left(\frac{8+\sqrt{34}}{8-\sqrt{34}}\right)

Therefore, the value of \int_{0}^{\frac{\pi}{2}}\frac{1}{5cosx+3sinx}dx is \frac{1}{\sqrt{34}}\log\left(\frac{8+\sqrt{34}}{8-\sqrt{34}}\right).

Question 5. \int_{0}^{a}\frac{x}{\sqrt{a^2+x^2}}dx

Solution:

We have,

I = \int_{0}^{a}\frac{x}{\sqrt{a^2+x^2}}dx

Let a² + x² = t². So, we have

=> 2x dx = 2t dt

=> x dx = t dt

Now, the lower limit is, x = 0

=> t² = a² + x²

=> t² = a² + 0²

=> t² = a²

=> t = a

Also, the upper limit is, x = a

=> t² = a² + x²

=> t² = a² + a²

=> t² = 2a²

=> t = √2 a

So, the equation becomes,

I = \int_{a}^{\sqrt{2}a}\frac{t}{t}dt

I = \int_{a}^{\sqrt{2}a}dt

I = \left[t\right]_{a}^{\sqrt{2}a}

I = √2a – a

I = a (√2 – 1)

Therefore, the value of \int_{0}^{a}\frac{x}{\sqrt{a^2+x^2}}dx is a (√2 – 1).

Question 6. \int_{0}^{1}\frac{e^x}{1+e^{2x}}dx

Solution:

We have,

I = \int_{0}^{1}\frac{e^x}{1+e^{2x}}dx

Let e^x = t. So, we have

=> e^x dx = dt

Now, the lower limit is, x = 0

=> t = e^x

=> t = e⁰

=> t = 1

Also, the upper limit is, x = a

=> t = e^x

=> t = e¹

=> t = e

So, the equation becomes,

I = \int_{1}^{e}\frac{1}{1+t^2}dt

I = \left[tan^{-1}t\right]_{1}^{e}

I = tan^{-1}e-tan^{-1}1

I = tan^{-1}e-\frac{\pi}{4}

Therefore, the value of \int_{0}^{1}\frac{e^x}{1+e^{2x}}dx is tan^{-1}e-\frac{\pi}{4}.

Question 7. \int_{0}^{1}xe^{x^2}dx

Solution:

We have,

I = \int_{0}^{1}xe^{x^2}dx

Let x² = t. So, we have

=> 2x dx = dt

Now, the lower limit is, x = 0

=> t = x²

=> t = 0²

=> t = 0

Also, the upper limit is, x = 1

=> t = x²

=> t = 1²

=> t = 1

So, the equation becomes,

I = \int_{0}^{1}\frac{e^t}{2}dt

I = \frac{1}{2}\int_{0}^{1}e^tdt

I = \frac{1}{2}\left[e^t\right]_{0}^{1}

I = \frac{1}{2}\left[e^1-e^0\right]

I = \frac{1}{2}(e-1)

Therefore, the value of \int_{0}^{1}xe^{x^2}dx is \frac{1}{2}(e-1).

Question 8. \int_{1}^{3}\frac{cos(logx)}{x}dx

Solution:

We have,

I = \int_{1}^{3}\frac{cos(logx)}{x}dx

Let log x = t. So, we have

=> (1/x) dx = dt

Now, the lower limit is, x = 1

=> t = log x

=> t = log 1

=> t = 0

Also, the upper limit is, x = 3

=> t = log x

=> t = log 3

So, the equation becomes,

I = \int_{0}^{\log3}costdt

I = \left[sint\right]_{0}^{\log3}

I = sin (log 3) – sin 0

I = sin (log 3) – 0

I = sin (log 3)

Therefore, the value of \int_{1}^{3}\frac{cos(logx)}{x}dx is sin (log 3).

Question 9. \int_{0}^{1}\frac{2x}{1+x^4}dx

Solution:

We have,

I = \int_{0}^{1}\frac{2x}{1+x^4}dx

Let x² = t. So, we have

=> 2x dx = dt

Now, the lower limit is, x = 0

=> t = x²

=> t = 0²

=> t = 0

Also, the upper limit is, x = 1

=> t = x²

=> t = 1²

=> t = 1

So, the equation becomes,

I = \int_{0}^{1}\frac{1}{1+t^2}dt

I = \left[tan^{-1}t\right]_{0}^{1}

I = tan^{-1}1-tan^{-1}0

I = \frac{\pi}{4}-0

I = \frac{\pi}{4}

Therefore, the value of \int_{0}^{1}\frac{2x}{1+x^4}dx is \frac{\pi}{4}.

Question 10. \int_{0}^{a}\sqrt{a^2-x^2}dx

Solution:

We have,

I = \int_{0}^{a}\sqrt{a^2-x^2}dx

Let x = a sin t. So, we have

=> dx = a cos t dt

Now, the lower limit is, x = 0

=> a sin t = x

=> a sin t = 0

=> sin t = 0

=> t = 0

Also, the upper limit is, x = a

=> a sin t = a

=> a sin t = a

=> sin t = 1

=> t = π/2

So, the equation becomes,

I = \int_{0}^{\frac{\pi}{2}}\sqrt{a^2-a^2sin^2t}(acost)dt

I = \int_{0}^{\frac{\pi}{2}}\sqrt{a^2(1-sin^2t)}(acost)dt

I = \int_{0}^{\frac{\pi}{2}}\sqrt{a^2cos^2t}(acost)dt

I = \int_{0}^{\frac{\pi}{2}}(acost)(acost)dt

I = \int_{0}^{\frac{\pi}{2}}a^2cos^2tdt

I = a^2\int_{0}^{\frac{\pi}{2}}cos^2tdt

I = \frac{a^2}{2}\int_{0}^{\frac{\pi}{2}}(1+cos2t)dt

I = \frac{a^2}{2}\left[t+\frac{sin2t}{2}\right]_{0}^{\frac{\pi}{2}}

I = \frac{a^2}{2}\left[\frac{\pi}{2}+0-0-0\right]

I = \frac{a^2}{2}\left[\frac{\pi}{2}\right]

I = \frac{\pi a^2}{4}

Therefore, the value of \int_{0}^{a}\sqrt{a^2-x^2}dx is \frac{\pi a^2}{4}.

Question 11. \int_{0}^{\frac{\pi}{2}}\sqrt{sinØ}cos^5ØdØ

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}}\sqrt{sinØ}cos^5ØdØ

I = \int_{0}^{\frac{\pi}{2}}\sqrt{sinØ}cos^4ØcosØdØ

Let sin Ø = t. So, we have

=> cos Ø dØ = dt

Now, the lower limit is, Ø = 0

=> t = sin Ø

=> t = sin 0

=> t = 0

Also, the upper limit is, Ø = π/2

=> t = sin Ø

=> t = sin π/2

=> t = 1

So, the equation becomes,

I = \int_{0}^{1}\sqrt{t}(1-t^2)^2dt

I = \int_{0}^{1}\sqrt{t}(1+t^4-2t^2)dt

I = \int_{0}^{1}(t^\frac{1}{2}+t^\frac{9}{2}-2t^\frac{5}{2})dt

I = \left[\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{t^{\frac{9}{2}+1}}{\frac{9}{2}+1}-\frac{2t^{\frac{5}{2}+1}}{\frac{5}{2}+1}\right]_{0}^{1}

I = \left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+\frac{t^{\frac{11}{2}}}{\frac{11}{2}}-\frac{2t^{\frac{7}{2}}}{\frac{7}{2}}\right]_{0}^{1}

I = \left[\frac{2t^{\frac{3}{2}}}{3}+\frac{2t^{\frac{11}{2}}}{11}-\frac{4t^{\frac{7}{2}}}{7}\right]_{0}^{1}

I = \frac{2}{3}+\frac{2}{11}-\frac{4}{7}

I = \frac{154+42-132}{231}

I = \frac{64}{231}

Therefore, the value of \int_{0}^{\frac{\pi}{2}}\sqrt{sinØ}cos^5ØdØ is \frac{64}{231}.

Question 12. \int_{0}^{\frac{\pi}{2}}\frac{cosx}{1+sinx}dx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}}\frac{cosx}{1+sin^2x}dx

Let sin x = t. So, we have

=> cos x dx = dt

Now, the lower limit is, x = 0

=> t = sin x

=> t = sin 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = sin x

=> t = sin π/2

=> t = 1

So, the equation becomes,

I = \int_{0}^{1}\frac{1}{1+t^2}dt

I = \left[tan^{-1}t\right]_{0}^{1}

I = tan^{-1}1-tan^{-1}0

I = \frac{\pi}{4}-0

I = \frac{\pi}{4}

Therefore, the value of \int_{0}^{\frac{\pi}{2}}\frac{cosx}{1+sin^2x}dx is \frac{\pi}{4}.

Question 13. \int_{0}^{\frac{\pi}{2}}\frac{sin\theta}{\sqrt{1+cos\theta}}d\theta

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}}\frac{sin\theta}{\sqrt{1+cos\theta}}d\theta

Let 1 + cos θ = t². So, we have

=> – sin θ dθ = 2t dt

=> sin θ dθ = –2t dt

Now, the lower limit is, θ = 0

=> t² = 1 + cos θ

=> t² = 1 + cos 0

=> t² = 1 + 1

=> t² = 2

=> t = √2

Also, the upper limit is, θ = π/2

=> t² = 1 + cos θ

=> t² = 1 + cos π/2

=> t² = 1 + 0

=> t² = 1

=> t = 1

So, the equation becomes,

I = \int_{\sqrt{2}}^{1}\frac{-2t}{t}dt

I = \int_{\sqrt{2}}^{1}-2dt

I = -2\left[t\right]_{\sqrt{2}}^{1}

I = -2\left[1-\sqrt{2}\right]

I = 2\left[\sqrt{2}-1\right]

Therefore, the value of \int_{0}^{\frac{\pi}{2}}\frac{sin\theta}{\sqrt{1+cos\theta}}d\theta is 2\left[\sqrt{2}-1\right].

Question 14. \int_{0}^{\frac{\pi}{3}}\frac{cosx}{3+4sinx}dx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{3}}\frac{cosx}{3+4sinx}dx

Let 3 + 4 sin x = t. So, we have

=> 0 + 4 cos x dx = dt

=> 4 cos x dx = dt

=> cos x dx = dt/4

Now, the lower limit is, x = 0

=> t = 3 + 4 sin x

=> t = 3 + 4 sin 0

=> t = 3 + 0

=> t = 3

Also, the upper limit is, x = π/3

=> t = 3 + 4 sin x

=> t = 3 + 4 sin π/3

=> t = 3 + 4 (√3/2)

=> t = 3 + 2√3

So, the equation becomes,

I = \int_{3}^{3+2\sqrt{3}}\frac{1}{4t}dt

I = \frac{1}{4}\left[\log t\right]_{3}^{3+2\sqrt{3}}

I = \frac{1}{4}\left[\log (3+2\sqrt{3})-\log 3)\right]

I = \frac{1}{4}\log\frac{3+2\sqrt{3}}{3}

Therefore, the value of \int_{0}^{\frac{\pi}{3}}\frac{cosx}{3+4sinx}dx is \frac{1}{4}\log\frac{3+2\sqrt{3}}{3}.

Question 15. \int_{0}^{1}\frac{\sqrt{tan^{-1}x}}{1+x^2}dx

Solution:

We have,

I = \int_{0}^{1}\frac{\sqrt{tan^{-1}x}}{1+x^2}dx

Let tan^–1 x = t. So, we have

=> (1/1+x²) dx = dt

Now, the lower limit is, x = 0

=> t = tan^–1 x

=> t = tan^–1 0

=> t = 0

Also, the upper limit is, x = 1

=> t = tan^–1 t

=> t = tan^–1 1

=> t = π/4

So, the equation becomes,

I = \int_{0}^{\frac{\pi}{4}}t^{\frac{1}{2}}dt

I = \left[\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{0}^{\frac{\pi}{4}}

I = \left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{\frac{\pi}{4}}

I = \frac{2}{3}\left[t^{\frac{3}{2}}\right]_{0}^{\frac{\pi}{4}}

I = \frac{2}{3}\left[(\frac{\pi}{4})^{\frac{3}{2}}-0\right]

I = \frac{2}{3}(\frac{\pi}{4})^{\frac{3}{2}}

I = \frac{2}{24}\pi^{\frac{3}{2}}

I = \frac{1}{12}\pi^{\frac{3}{2}}

Therefore, the value of \int_{0}^{1}\frac{\sqrt{tan^{-1}x}}{1+x^2}dx is \frac{1}{12}\pi^{\frac{3}{2}}.

Question 16. \int_{0}^{2}x\sqrt{x+2}dx

Solution:

We have,

I = \int_{0}^{2}x\sqrt{x+2}dx

Let x + 2 = t². So, we have

=> dx = 2t dt

Now, the lower limit is, x = 0

=> t² = x + 2

=> t² = 0 + 2

=> t² = 2

=> t = √2

Also, the upper limit is, x = 2

=> t² = x + 2

=> t² = 2 + 2

=> t² = 4

=> t = 2

So, the equation becomes,

I = \int_{\sqrt{2}}^{2}(t^2-2)\sqrt{t^2}2tdt

I = 2\int_{\sqrt{2}}^{2}(t^2-2)t^2dt

I = 2\int_{\sqrt{2}}^{2}(t^4-2t^2)dt

I = 2\left[\frac{t^5}{5}-\frac{2t^3}{3}\right]_{\sqrt{2}}^{2}

I = 2\left[\frac{32}{5}-\frac{16}{3}-\frac{4\sqrt{2}}{5}+\frac{4\sqrt{2}}{3}\right]

I = 2\left[\frac{16+8\sqrt{2}}{15}\right]

I = 2\left[\frac{8\sqrt{2}(1+\sqrt{2})}{15}\right]

I = \frac{16\sqrt{2}(1+\sqrt{2})}{15}

Therefore, the value of \int_{0}^{2}x\sqrt{x+2}dx is \frac{16\sqrt{2}(1+\sqrt{2})}{15}.

Question 17. \int_{0}^{1}tan^{-1}\frac{2x}{1-x^2}dx

Solution:

We have,

I = \int_{0}^{1}tan^{-1}\frac{2x}{1-x^2}dx

Let x = tan t. So, we have

=> dx = sec² t dt

Now, the lower limit is, x = 0

=> tan t = x

=> tan x = 0

=> x = 0

Also, the upper limit is, x = 1

=> tan t = x

=> tan x = 1

=> x = π/4

So, the equation becomes,

I = \int_{0}^{\frac{\pi}{4}}tan^{-1}(\frac{2tant}{1-tan^2t})sec^2tdt

I = \int_{0}^{\frac{\pi}{4}}tan^{-1}(tan2t)sec^2tdt

I = \int_{0}^{\frac{\pi}{4}}2tsec^2tdt

I = 2\int_{0}^{\frac{\pi}{4}}tsec^2tdt

On applying integration by parts method, we get

I = 2\left[t\int_0^\frac{\pi}{4}sec^2tdt-\int_0^\frac{\pi}{4}tantdt\right]

I = 2\left[ttant-log(cost)\right]^{\frac{\pi}{4}}_0

I = 2\left[\frac{\pi}{4}+log\frac{1}{\sqrt{2}}-0-0\right]

I = 2\left[\frac{\pi}{4}+\frac{1}{2}log2\right]

I = \frac{\pi}{2}+log2

Therefore, the value of \int_{0}^{1}tan^{-1}\frac{2x}{1-x^2}dx is \frac{\pi}{2}+log 2.

Question 18. \int_{0}^{\frac{\pi}{2}}\frac{sinxcosx}{1+sin^4x}dx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}}\frac{sinxcosx}{1+sin^4x}dx

Let sin² x = t. So, we have

=> 2 sin x cos x = dt

=> sin x cos x = dt/2

Now, the lower limit is, x = 0

=> t = sin² x

=> t = sin² 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = sin² x

=> t = sin² π/2

=> t = 1

So, the equation becomes,

I = \int_{0}^{1}\frac{1}{2(1+t^2)}dt

I = \frac{1}{2}\int_{0}^{1}\frac{1}{1+t^2}dt

I = \frac{1}{2}\left[tan^{-1}t\right]^1_0

I = \frac{1}{2}\left[tan^{-1}1-tan^{-1}0\right]

I = \frac{1}{2}\left[\frac{\pi}{4}-0\right]

I = \frac{1}{2}(\frac{\pi}{4})

I = \frac{\pi}{8}

Therefore, the value of \int_{0}^{\frac{\pi}{2}}\frac{sinxcosx}{1+sin^4x}dx is \frac{\pi}{8}.

Question 19. \int_{0}^{\frac{\pi}{2}}\frac{1}{acosx+bsinx}dx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}}\frac{1}{acosx+bsinx}dx

On putting cos x = \frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}=\frac{1-tan^2\frac{x}{2}}{sec^2\frac{x}{2}}     and sin x = \frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}=\frac{2tan\frac{x}{2}}{sec^2\frac{x}{2}}    , we get,

I = \int_{0}^{\frac{\pi}{2}}\frac{sec^2\frac{x}{2}}{a(1-tan^2\frac{x}{2})+2btan\frac{x}{2}}dx

Let tan x/2 = t. So, we have

=> 1/2 sec² x/2 dx = dt

=> sec² x/2 dx = 2 dt

Now, the lower limit is, x = 0

=> t = tan x/2

=> t = tan 0/2

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = tan x/2

=> t = tan π/4

=> t = 1

So, the equation becomes,

I = 2\int_{0}^{1}\frac{1}{a(1-t^2)+2bt}dt

I = 2\int_{0}^{1}\frac{1}{a-at^2+2bt}dt

I = 2\int_{0}^{1}\frac{1}{-a(t^2-\frac{2b}{a}t-1)}dt

I = \frac{2}{a}\int_{0}^{1}\frac{1}{-\left[(t-\frac{b}{a})^2-1-\frac{b^2}{a^2}\right]}dt

I = \frac{2}{a}\int_{0}^{1}\frac{1}{(\frac{b^2}{a^2}+1)-(t-\frac{b}{a})^2}dt

I = \left[\frac{2}{a}\left(\frac{1}{2\sqrt{\frac{b^2+a^2}{a^2}}}\right)\log\left(\frac{\sqrt{\frac{b^2+a^2}{a^2}}+(t-\frac{b}{a})}{\sqrt{\frac{b^2+a^2}{a^2}}-(t-\frac{b}{a})}\right)\right]_{0}^{1}

I = \frac{1}{\sqrt{b^2+a^2}}\log\left(\frac{a+b+\sqrt{a^2+b^2}}{a+b-\sqrt{a^2+b^2}}\right)

Therefore, the value of \int_{0}^{\frac{\pi}{2}}\frac{1}{acosx+bsinx}dx is \frac{1}{\sqrt{b^2+a^2}}\log\left(\frac{a+b+\sqrt{a^2+b^2}}{a+b-\sqrt{a^2+b^2}}\right).

Question 20. \int_{0}^{\frac{\pi}{2}}\frac{1}{5+4sinx}dx

Solution:

We have,

I = \int_{0}^{\frac{\pi}{2}}\frac{1}{5+4sinx}dx

On putting sin x = \frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}, we get

I = \int_{0}^{\frac{\pi}{2}}\frac{1}{5+4\left(\frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}\right)}dx

I = \int_{0}^{\frac{\pi}{2}}\frac{1}{5+\left(\frac{8tan\frac{x}{2}}{1+tan^2\frac{x}{2}}\right)}dx

I = \int_{0}^{\frac{\pi}{2}}\frac{1}{\frac{5(1+tan^2\frac{x}{2})+8tan\frac{x}{2}}{1+tan^2\frac{x}{2}}}dx

I = \int_{0}^{\frac{\pi}{2}}\frac{1}{\frac{5+5tan^2\frac{x}{2}+8tan\frac{x}{2}}{1+tan^2\frac{x}{2}}}dx

I = \int_{0}^{\frac{\pi}{2}}\frac{1}{\frac{5+5tan^2\frac{x}{2}+8tan\frac{x}{2}}{sec^2\frac{x}{2}}}dx

I = \int_{0}^{\frac{\pi}{2}}\frac{sec^2\frac{x}{2}}{5+5tan^2\frac{x}{2}+8tan\frac{x}{2}}dx

Let tan x/2 = t. So, we have

=> 1/2 sec² x/2 dx = dt

=> sec² x/2 dx = 2 dt

Now, the lower limit is, x = 0

=> t = tan x/2

=> t = tan 0/2

=> t = tan 0

=> t = 0

Also, the upper limit is, x = π/2

=> t = tan x/2

=> t = tan π/4

=> t = 1

So, the equation becomes,

I = \int_{0}^{1}\frac{1}{5+5t^2+8t}dt

I = \frac{2}{5}\int_{0}^{1}\frac{1}{1+t^2+\frac{8}{5}t}dt

I = \frac{2}{5}\int_{0}^{1}\frac{1}{1-\frac{16}{25}+\frac{16}{25}+t^2+\frac{8}{5}t}dt

I = \frac{2}{5}\int_{0}^{1}\frac{1}{\frac{9}{25}+\frac{16}{25}+t^2+\frac{8}{5}t}dt

I = \frac{2}{5}\int_{0}^{1}\frac{1}{(\frac{3}{5})^2+(t+\frac{4}{5})^2}dt

I = \frac{2}{5}\left[\frac{5}{3}tan^{-1}\frac{5}{3}(t+\frac{4}{5})\right]_{0}^{1}

I = \frac{2}{3}\left[tan^{-1}(\frac{5t}{3}+\frac{4}{3})\right]_{0}^{1}

I = \frac{2}{3}\left[tan^{-1}(\frac{5}{3}+\frac{4}{3})-tan^{-1}(0+\frac{4}{3})\right]

I = \frac{2}{3}\left[tan^{-1}(\frac{9}{3})-tan^{-1}(\frac{4}{3})\right]

I = \frac{2}{3}\left[tan^{-1}(3)-tan^{-1}(\frac{4}{3})\right]

I = \frac{2}{3}\left[tan^{-1}(\frac{3-\frac{4}{3}}{1+3(\frac{4}{3})})\right]

I = \frac{2}{3}\left[tan^{-1}(\frac{\frac{5}{3}}{1+4})\right]

I = \frac{2}{3}tan^{-1}(\frac{\frac{5}{3}}{5})

I = \frac{2}{3}tan^{-1}(\frac{1}{3})

Therefore, the value of \int_{0}^{\frac{\pi}{2}}\frac{1}{5+4sinx}dx is \frac{2}{3}tan^{-1}(\frac{1}{3}).

Summary

This exercise typically covers:

• Basic evaluation of definite integrals
• Application of fundamental theorem of calculus
• Integrals involving algebraic and trigonometric functions
• Use of substitution method in definite integrals
• Properties of definite integrals