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  • Need solution for rd sharma maths class 12 chapter Indefinite Integrals exercise 18.2 question 21

Need solution for rd sharma maths class 12 chapter Indefinite Integrals exercise 18.2 question 21

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x-sin x +c

Hint:To solve this equation we replace sin^{2}x by 1-cos^{2}x

\begin{aligned} &\text { Given: } \int \frac{\sin ^{2} x}{1+\cos ^{2} x} d x \\ &\text { Solution: } \frac{\sin ^{2} x}{1+\cos x} \end{aligned}

\begin{aligned} &=\frac{1-\cos ^{2} x}{1+\cos x}\left[\begin{array}{l} \operatorname{Sin}^{2} x+\cos ^{2} x=1 \\ \operatorname{Sin}^{2} x=1-\cos ^{2} x \end{array}\right] \\ &=\frac{\left(1-\cos x\right)\left(1+\cos x\right)}{1+\cos x} \quad\left[a^{2}-b^{2}=(a+b)(a-b)\right] \\ &=1-\cos x \\ &\int \frac{\sin ^{2} x}{1+\cos x} d x=\int 1-\cos x d x \\ &=x-\sin x+c \end{aligned}

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