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

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

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

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$\displaystyle \int \frac{1\, -\, x^7}{x(1\, +\, x^7)} dx$ equals
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