Definite Integration Question 36
Question 36
- The value of $\int_{0}^{\pi / 2} \frac{d x}{1+\tan ^{3} x}$ is
(1993, 1M) (a) 0 (b) 1 (c) $\pi / 2$ (d) $\pi / 4$
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Solution:
- Let $I=\int_{0}^{\pi / 2} \frac{1}{1+\tan ^{3} x} d x=\int_{0}^{\pi / 2} \frac{1}{1+\frac{\sin ^{3} x}{\cos ^{3} x}} d x$
$$ \Rightarrow \quad I=\int_{0}^{\pi / 2} \frac{\cos ^{3} x}{\cos ^{3} x+\sin ^{3} x} d x $$
$\Rightarrow \quad I=\int_{0}^{\pi / 2} \frac{\cos ^{3} \frac{\pi}{2}-x}{\cos ^{3} \frac{\pi}{2}-x+\sin ^{3} \frac{\pi}{2}-x} d x$
$\Rightarrow \quad I=\int_{0}^{\pi / 2} \frac{\sin ^{3} x}{\sin ^{3} x+\cos ^{3} x} d x$
On adding Eqs. (i) and (ii), we get
$$ 2 I=\int_{0}^{\pi / 2} 1 d x \Rightarrow 2 I=[x]_{0}^{\pi / 2}=\pi / 2 \Rightarrow I=\pi / 4 $$
