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  • Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 62 maths.

Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 62 maths.

Answers (1)

Answer :   \frac{2}{2-n}\left [ 2^{1-\frac{n}{2}}-1 \right ]

Hint: Use indefinite formula and the given limits to solve this integral
Given:  \int_{0}^{\pi / 2} \frac{\cos x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{n}} d x

Solution: \int_{0}^{\pi / 2} \frac{\cos x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{n}} d x

\begin{aligned} &=\int_{0}^{\pi / 2} \frac{\cos ^{2} \frac{x}{2}- \sin ^{2} \frac{x}{2}}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{n}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \cos 2 \theta=\cos ^{2} \theta-\sin ^{2} \theta\right] \\ &=\int_{0}^{\pi / 2} \frac{\left(\cos _{2}^{x}-\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{n}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad \quad\left[\left(\boldsymbol{a}^{2}-b^{2}\right)=(a+b)(a-b)\right] \end{aligned}

=\int_{0}^{\pi / 2} \frac{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{n-1}} d x

 

Put  \cos \frac{x}{2}+\sin \frac{x}{2}=t \Rightarrow\left(-\sin \frac{x}{2} \cdot \frac{1}{2}+\cos \frac{x}{2} \cdot \frac{1}{2}\right) d x=d t

\begin{aligned} &\Rightarrow \frac{1}{2}\left[\cos \frac{x}{2}-\sin \frac{x}{2}\right] d x=d t \\ &\Rightarrow\left[\cos \frac{x}{2}-\sin \frac{x}{2}\right] d x=2 d t \end{aligned}

when x=\frac{\pi}{2}\Rightarrow t =\sqrt{2}

\begin{aligned} &\therefore \int_{0}^{\pi / 2} \frac{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{n-1}} d x \\ &=\int_{1}^{\sqrt{2}} \frac{1}{t^{n-1}} 2 d t \\ &=2 \int_{1}^{\sqrt{2}} t^{-(n-1)} d t \\ &=2 \int_{1}^{\sqrt{2}} t^{-n+1} d t \end{aligned}

\begin{aligned} &=2\left[\frac{t^{-n+1+1}}{-n+1+1}\right]_{1}^{\sqrt{2}} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\therefore \int x^{n} d x=\frac{x^{n+1}}{n+1}\right] \\ &=2\left[\frac{t^{-n+2}}{-n+2}\right]_{1}^{\sqrt{2}} \\ &=\frac{2}{(2-n)}\left[\frac{1}{t^{n-2}}\right]_{1}^{\sqrt{2}} \\ &=\frac{2}{(2-n)}\left[\frac{1}{(\sqrt{2})^{n-2}}-\frac{1}{1^{n-2}}\right] \end{aligned}

\begin{aligned} &=\frac{2}{2-n}\left[\frac{1}{2\left(\frac{n-2}{2}\right)}-1\right] \\ &=\frac{2}{2-n}\left[\frac{1}{2\left(\frac{n}{2}-1\right)}-1\right] \\ &=\frac{2}{2-n}\left[2^{1-\frac{n}{2}}-1\right] \end{aligned}

 

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