Indefinite Integration 1 Question 5
6. The value of $\int _0^{\pi / 2} \frac{\sin ^{3} x}{\sin x+\cos x} d x$ is
(a) $\frac{\pi-1}{2}$
(b) $\frac{\pi-2}{8}$
(c) $\frac{\pi-1}{4}$
(d) $\frac{\pi-2}{4}$
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Solution:
Key Idea Use property of definite integral.
$\int _a^{b} f(x) d x=\int _a^{b} f(a+b-x) d x$
Let
$$ I=\int _0^{\frac{\pi}{2}} \frac{\sin ^{3} x}{\sin x+\cos x} d x $$
On applying the property, $\int _a^{b} f(x) d x=\int _a^{b} f(a+b-x) d x$, we get
$$ I=\int _0^{\pi / 2} \frac{\cos ^{3} x}{\cos x+\sin x} d x $$
On adding integrals (i) and (ii), we get
$$ \begin{aligned} 2 I & =\int _0^{\pi / 2} \frac{\sin ^{3} x+\cos ^{3} x}{\sin x+\cos x} d x \\ & =\int _0^{\frac{\pi}{2}} \frac{(\sin x+\cos x)\left(\sin ^{2} x+\cos ^{2} x-\sin x \cos x\right)}{\sin x+\cos x} d x \\ & =\int _0^{\frac{\pi}{2}} 1-\frac{1}{2}(2 \sin x \cos x) d x \end{aligned} $$
$=\int _0^{\frac{\pi}{2}} 1-\frac{1}{2} \sin 2 x d x$
$$ \begin{aligned} & =x+\frac{1}{4} \cos 2 x _0^{\pi / 2}=\frac{\pi}{2}-0+\frac{1}{4}(-1-1)=\frac{\pi}{2}-\frac{1}{2} \\ \Rightarrow \quad I & =\frac{\pi}{4}-\frac{1}{4}=\frac{\pi-1}{4} \end{aligned} $$
