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$\int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}$ is equal to :
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Verified Answer
The correct answer is:
$\frac{\pi}{4}$
$I=\int_0^{\pi / 2} \frac{d x}{1+\tan ^3 x}$
$I=\int_0^{\pi / 2} \frac{\cos ^3 x}{\sin ^3 x+\cos ^3 x} d x$ ...(i)
$I=\int_0^{\pi / 2} \frac{\cos ^3\left(\frac{\pi}{2}-x\right)}{\sin ^3\left(\frac{\pi}{2}-x\right)+\cos ^3\left(\frac{\pi}{2}-x\right)} d x$
$I=\int_0^{\pi / 2} \frac{\sin ^3 x}{\sin ^3 x+\cos ^3 x} d x$ ...(ii)
On adding Eqs. (i) and (ii), we get
$2 I=\int_0^{\pi / 2} \frac{\sin ^3 x+\cos ^3 x}{\sin ^3 x+\cos ^3 x} d x$
$=\int_0^{\pi / 2} 1 d x=\frac{\pi}{2}$
$\Rightarrow I=\frac{\pi}{4}$
$I=\int_0^{\pi / 2} \frac{\cos ^3 x}{\sin ^3 x+\cos ^3 x} d x$ ...(i)
$I=\int_0^{\pi / 2} \frac{\cos ^3\left(\frac{\pi}{2}-x\right)}{\sin ^3\left(\frac{\pi}{2}-x\right)+\cos ^3\left(\frac{\pi}{2}-x\right)} d x$
$I=\int_0^{\pi / 2} \frac{\sin ^3 x}{\sin ^3 x+\cos ^3 x} d x$ ...(ii)
On adding Eqs. (i) and (ii), we get
$2 I=\int_0^{\pi / 2} \frac{\sin ^3 x+\cos ^3 x}{\sin ^3 x+\cos ^3 x} d x$
$=\int_0^{\pi / 2} 1 d x=\frac{\pi}{2}$
$\Rightarrow I=\frac{\pi}{4}$
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