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KEAM 2013 Math Paper

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Question : 116 of 120

Marks: +1, -0

\int\limits_{0}^{\pi/2} \frac{dx}{1+\tan^{3}x}=

1
$\pi$
$\pi/2$
$\pi/4$
0

Solution:

Let

I=\int\limits_{0}^{\pi/2} \;\frac{dx}{1+\tan^{3}x}

\Rightarrow I=\int\limits_{0}^{\pi/2} \frac{\cos^{3}x}{\sin^{3}x+\cos^{3}x} dx

.....(i)

\Rightarrow I=\int\limits_{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)} dx

\{ \because \int\limits_{0}^{a} f(x)=\int\limits_{0}^{a} f(a-x) dx \}

\Rightarrow I=\int\limits_{0}^{\pi/2} \frac{\sin^{3}x}{\cos^{3}x+\sin^{3}x} dx

....(ii)

On adding Eqs. (i) and (ii), we get

2l=\int\limits_{0}^{\pi/2} \frac{\sin^{3}x+\cos^{3}x}{\sin^{3}x+\cos^{3}x} dx

=\int\limits_{0}^{\pi/2} (1) dx = [x]_{0}^{\pi/2}

\Rightarrow \;\; 2I = \frac{\pi}{2} \Rightarrow I = \frac{\pi}{4}

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