Mathematics

Solve $$\displaystyle\int\dfrac{1}{{\sin x{{\cos }^3}x}}dx$$


SOLUTION

Consider the given integral.

$$I=\int{\dfrac{dx}{\sin x{{\cos }^{3}}x}}$$

$$ I=\int{\dfrac{\sin xdx}{{{\sin }^{2}}x{{\cos }^{3}}x}} $$

$$ I=\int{\dfrac{\sin xdx}{\left( 1-{{\cos }^{2}}x \right){{\cos }^{3}}x}} $$

 

Let $$t=\cos x$$

$$ \dfrac{dt}{dx}=-\sin x $$

$$ -dt=\sin xdx $$

 

Therefore,

$$ I=-\int{\dfrac{dt}{\left( 1-{{t}^{2}} \right){{t}^{3}}}} $$

$$ I=\int{\dfrac{dt}{\left( {{t}^{2}}-1 \right){{t}^{3}}}} $$

$$ I=\int{\dfrac{dt}{\left( t+1 \right)\left( t-1 \right){{t}^{3}}}} $$

$$ I=\int{\dfrac{dt}{2\left( t+1 \right)}}+\int{\dfrac{dt}{2\left( t-1 \right)}}-\int{\dfrac{dt}{t}}-\int{\dfrac{dt}{{{t}^{3}}}} $$

$$ I=\dfrac{1}{2}\ln \left( t+1 \right)+\dfrac{1}{2}\ln \left( t-1 \right)-\ln \left( t \right)+\dfrac{1}{2{{t}^{2}}}+C $$

 

On putting the value of $$t$$, we get

$$I=\dfrac{1}{2}\ln \left( \cos x+1 \right)+\dfrac{1}{2}\ln \left( \cos x-1 \right)-\ln \left( \cos x \right)+\dfrac{1}{2{{\left( \cos x \right)}^{2}}}+C$$

 

Hence, this is the answer.

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Subjective Medium Published on 17th 09, 2020
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