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  • Evaluate the definite integrals in Exercises 25 to 33. 26

Evaluate the definite integrals in Exercises 25 to 33.

    Q26.  \int_0^\frac{\pi}{4}\frac{\sin x\cos x }{\cos^4 x+\sin^4 x}

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\int_0^\frac{\pi}{4}\frac{\sin x\cos x }{\cos^4 x+\sin^4 x}

First, let's convert sin and cos into tan and sec. (because we have a good relation in tan and square of sec,)

Let's divide both the numerator and denominator by cos^4x

=\int_0^\frac{\pi}{4}\frac{\frac{\sin x\cos x}{cosxcosxsos^2x} }{1+\frac{\sin^4 x}{\cos^4x}}

=\int_0^\frac{\pi}{4}\frac{tanxsec^2x}{1+tan^4x}

Now let's change the variable

\\t=tan^2x \\dt=2tanxsec^2xdx

the limits will also change since the variable is changing 

when\:x=0,t=tan^20=0

when\:x=\frac{\pi}{4},t=tan^2\frac{\pi}{4}=1

So, the integration becomes:

I=\frac{1}{2}\int_{0}^{1}\frac{dt}{1+t^2}

I=\frac{1}{2}\left [ tan^{-1}t \right ]_0^1

I=\frac{1}{2}\left [ tan^{-1}1 \right ]-\frac{1}{2}\left [ tan^{-1}0\right ]

I=\frac{1}{2}\left [ \frac{\pi}{4} \right ]-0

I=\frac{\pi}{8}

Posted by

Pankaj Sanodiya

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