Test Index

AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

Section: Mathematics

© examsnet.com

Question : 77 of 160

Marks: +1, -0

\int\limits_{0}^{\frac{\pi}{2}} \frac{\sin^{3} x \cos x \, dx}{\sin^{4} x + \cos^{4} x}

=

π
$\frac{\pi}{2}$
$\frac{\pi}{4}$
$\frac{\pi}{8}$

Solution:

Let's consider the given expression as

I = \int\limits_{0}^{\;\frac{\pi}{2}} \;\frac{\sin^{3} x \cos x \, dx}{\sin^{4} x + \cos^{4} x} \;\;\;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot (I)

Then

I \;\; = \int\limits_{0}^{\;\frac{\pi}{2}} \;\frac{\sin^{3}\left(\frac{\pi}{2} - x\right) \cos\left(\frac{\pi}{2} - x\right)}{\sin^{4}\left(\;\frac{\pi}{2} - x\right) + \cos^{4}\left(\;\frac{\pi}{2} - x\right)} \, dx

\;\; = \int\limits_{0}^{\;\frac{\pi}{2}} \;\frac{\cos^{3} x \sin x}{\cos^{4} x + \sin^{4} x} \, dx

Add equation (I) and (II),

2I \;\; = \int\limits_{0}^{\;\frac{\pi}{2}} \;\frac{\sin^{3} x \cos x + \cos^{3} x \sin x}{\cos^{4} x + \sin^{4} x} \, dx

\;\; = \int\limits_{0}^{\;\frac{\pi}{2}} \;\frac{\sin x \cos x \left(\sin^{2} x + \cos^{2} x\right)}{\cos^{4} x + \sin^{4} x} \, dx

\;\; = \int\limits_{0}^{\;\frac{\pi}{2}} \;\frac{\sin x \cos x}{\cos^{4} x + \sin^{4} x} \, dx

\;\; = \int\limits_{0}^{\;\frac{\pi}{2}} \;\frac{\sin x \cos x}{\cos^{4} x \left(\tan^{4} x + 1\right)} \, dx

Further simplify the above expression,

2I = \int\limits_{0}^{\;\frac{\pi}{2}} \;\frac{\tan x \sec^{2} x}{\left(\tan^{4} x + 1\right)} \, dx

Let

\tan^{2} x = t

then

2 \tan x \sec^{2} x \, dx = dt

Now

2I \;\; = \;\frac{1}{2} \int\limits_{0}^{\infty} \;\frac{dt}{t + 1}

\;\; = \;\frac{1}{2} \left[\tan^{-1} t\right]_{0}^{\infty}

\;\; = \;\frac{1}{2} \left[\tan^{-1}(\infty) - \tan^{-1}(0)\right]

\;\; = \;\frac{1}{2} \times \;\frac{\pi}{2}

Thus,

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

I = \;\frac{\pi}{8}

© examsnet.com

Go to Question:

Prev Question

Next Question