Question

Integrate 1/sin x cos^3x wrt x

Answer 1

multiply and divide by cos square x and in numerator you will get sec square x anding denominator you will get tanx

Put tanx =t
Sec square xdx = dt

Integration will be Logt+c
Log(tanx) +c

Hope you get your answer.

Answer 2

$\frac{1}{2}\sec^{2}x + ln |\csc 2x - \cot 2x| + c$

Step-by-step explanation:

We have to evaluate the integration of $\frac{1}{\sin x \cos^{3}x }$

Now, $\int\ {\frac{1}{\sin x \cos^{3}x } } \, dx$

= $\int\ {\frac{\sin^{2}x + \cos^{2}x}{\sin x \cos^{3} x } } \, dx$ {Since, $\sin^{2}x + \cos^{2}x = 1$}

= $\int\ {\frac{\sin x}{\cos^{3}x } } \, dx + \int\ {\frac{1}{\sin x \cos x } } \, dx$

= $-\int\ {\frac{1}{\cos^{3}x } } \, d(\cos x) + 2\int\ {\frac{1}{\sin 2x} } \, dx$

= $\frac{\cos^{-2}x }{2} + 2\int\ {\csc 2x} \, dx + c$ {Where c is an integration constant}

= $\frac{1}{2}\sec^{2}x + \int\ {\csc 2x} \, d(2x)$

= $\frac{1}{2}\sec^{2}x + ln |\csc 2x - \cot 2x| + c$ (Answer)