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  • Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 57 maths.

Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 57 maths.

Answers (1)

Answer :   \left ( \frac{\log m}{m^2-1} \right )

Hint: Use indefinite integral formula and the limits to solve this integral

Given: \int_{0}^{\frac{\pi}{2}}\frac{\tan x}{1+m^2\tan^2x}dx

Solution: \int_{0}^{\frac{\pi}{2}}\frac{\tan x}{1+m^2\tan^2x}dx

=\int_{0}^{\frac{\pi}{2}} \frac{\frac{\sin x}{\cos x}}{1+m^{2} \frac{\sin ^{2} x}{\cos ^{2} x}} d x

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \frac{\frac{\sin x}{\cos x}}{\frac{\cos ^{2} x+m^{2} \sin ^{2} x}{\cos ^{2} x}} d x \\ &=\int_{0}^{\frac{\pi}{2}}\left(\frac{\sin x}{\cos x} \frac{\cos ^{2} x}{\cos ^{2} x+m^{2} \sin ^{2} x}\right) d x \\ &=\int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{\cos ^{2} x+m^{2} \sin ^{2} x} d x \end{aligned}

put  sin^2x=t \Rightarrow 2\sin x\cos x dx=dt \sin x \cos x=\frac{dt}{2}

when x=0 then t=0 and x=\frac{\pi}{2}  then  t=1

Therefore ,

 \begin{aligned} &\int_{0}^{\frac{\pi}{2}} \frac{\tan x}{1+m^{2} \tan ^{2} x} d x \\ &=\int_{0}^{1} \frac{1}{\left(1-\sin ^{2} x\right)+m^{2} t} \frac{d t}{2} \\ &=\frac{1}{2} \int_{0}^{1} \frac{1}{(1-t)+m^{2} t} d t \\ &=\frac{1}{2} \int_{0}^{1} \frac{1}{\left(m^{2}-1\right) t+1} d t \\ &\text { put } u=\left(m^{2}-1\right) t+1 \Rightarrow d u=\left(m^{2}-1\right) d t \Rightarrow d t=\frac{1}{m^{2}-1} d u \end{aligned}

\begin{aligned} &=\frac{1}{2} \int_{0}^{1} \frac{1}{u} \frac{1}{\left(m^{2}-1\right)} d u=\frac{1}{2\left(m^{2}-1\right)} \int_{0}^{1} \frac{1}{u} d u \\ &=\frac{1}{2\left(m^{2}-1\right)}[\log |u|]_{0}^{1} \\ &=\frac{1}{2\left(m^{2}-1\right)}\left[\log \left|\left(m^{2}-1\right) t+1\right|\right]_{0}^{1} \\ &=\frac{1}{2\left(m^{2}-1\right)}\left[\log \left|\left(m^{2}-1\right) 1+1\right|-\log \left|\left(m^{2}-1\right) 0+1\right|\right] \end{aligned}

\begin{aligned} &=\frac{1}{2\left(m^{2}-1\right)}\left[\log \left|\left(m^{2}-1\right) 1+1\right|-\log \left|\left(m^{2}-1\right) 0+1\right|\right] \\ &=\frac{1}{2\left(m^{2}-1\right)}\left[\log \left(m^{2}-1+1\right)-\log 1\right] \\ &=\frac{1}{2\left(m^{2}-1\right)}\left[\log m^{2}-0\right] \\ &=\frac{1}{2\left(m^{2}-1\right)} \log m^{2} \\ &=\frac{2 \log m}{2\left(m^{2}-1\right)} \\ &=\frac{\log m}{\left(m^{2}-1\right)} \end{aligned}

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