Mathematics

# Solve $\displaystyle\int\dfrac{{{{\sin }^4}x}}{{{{\cos }^2}x}}dx$

##### SOLUTION

We have,

$I=\displaystyle\int{\dfrac{{{\sin }^{4}}x}{{{\cos }^{2}}x}dx}$

$I=\displaystyle\int{\dfrac{{{\left( 1-{{\cos }^{2}}x \right)}^{2}}}{{{\cos }^{2}}x}dx}$

$I=\displaystyle\int{\dfrac{\left( 1+{{\cos }^{4}}x-2{{\cos }^{2}}x \right)}{{{\cos }^{2}}x}dx}$

$I=\displaystyle\int{\left( \dfrac{1}{{{\cos }^{2}}x}+{{\cos }^{2}}x-2 \right)dx}$

$I=\displaystyle\int{\left( {{\sec }^{2}}x+{{\cos }^{2}}x-2 \right)dx}$

We know that

${{\cos }^{2}}x=\dfrac{1+\cos 2x}{2}$

Therefore,

$I=\displaystyle\int{\left( {{\sec }^{2}}x+\dfrac{1+\cos 2x}{2}-2 \right)dx}$

$I=\tan x+\dfrac{1}{2}\left( x+\dfrac{\sin 2x}{2} \right)-2x+C$

Hence, this is the answer.

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Subjective Medium Published on 17th 09, 2020
Questions 203525
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