Question

Evaluate the integral $\int\dfrac{ln\left(1+sin^2x\right)}{cos^2x}dx$.

Answer 1

Hint: To evaluate such complex integrals we need to be aware about several formulae related to integration. Since, this is not a standard form of any integration; we will take the cosine inverse and convert it into secant inverse. After we get two such functions, we will use the by parts integration. We will use the standard ILATE rule to choose the first and second function.

Complete step-by-step solution:
We have,
$\int\dfrac{ln\left(1+sin^2x\right)}{cos^2x}dx$
We can write this as,
$\int{\left( {{\sec }^{2}}x\times ln\left( 1+si{{n}^{2}}x \right) \right)}dx$
Now, we apply by parts by taking $ln\left( 1+si{{n}^{2}}x \right)$ as the first function and ${{\sec }^{2}}x$ as the second function.
So, we get:
\[ln\left( 1+si{{n}^{2}}x \right)\int{{{\sec }^{2}}x-\int{\int{{{\sec }^{2}}x}\times \dfrac{d\left( ln\left( 1+si{{n}^{2}}x \right) \right)}{dx}}}dx\]
\[\begin{align}
  & =ln\left( 1+si{{n}^{2}}x \right)\tan x-\int{\tan x}\times \dfrac{2\sin x\cos x}{1+{{\sin }^{2}}x}dx \\
 & =\tan x\ln \left( 1+{{\sin }^{2}}x \right)-2\int{\frac{{{\sin }^{2}}x}{1+{{\sin }^{2}}x}dx} \\
 & =\tan x\ln \left( 1+{{\sin }^{2}}x \right)-2\int{\left( 1-\frac{1}{1+{{\sin }^{2}}x} \right)dx} \\
\end{align}\]
Now dividing numerator and denominator by $cos^2x$ in the last term:
\[\begin{align}
  & \tan x\ln \left( 1+{{\sin }^{2}}x \right)-2\int{\left( 1-\dfrac{{{\sec }^{2}}x}{2{{\tan }^{2}}x+1} \right)dx} \\
 & =\tan x\ln \left( 1+{{\sin }^{2}}x \right)-2x+2\int{\dfrac{dt}{2{{t}^{2}}+1}} \\
\end{align}\]
We have done the following substitution in the last step:
Let $t=\tan x$
Then $dt=sec^2xdx$
Now, we use the formula:
\[\int{\dfrac{dt}{{{t}^{2}}+{{a}^{2}}}=\frac{1}{a}{{\tan }^{-1}}\left( \frac{x}{a} \right)}\]
Using this we obtain:
\[\tan x\ln \left( 1+{{\sin }^{2}}x \right)-2x+2\left( \sqrt{2} \right){{\tan }^{-1}}t\sqrt{2}+c\]
Now putting back the value of $t$ we have:
\[\tan x\ln \left( 1+{{\sin }^{2}}x \right)-2x+2\sqrt{2}{{\tan }^{-1}}\left( \sqrt{2}\tan x \right)+c\]
Hence,
$\int{\dfrac{ln\left( 1+si{{n}^{2}}x \right)}{co{{s}^{2}}x}}dx=\tan x\ln \left( 1+{{\sin }^{2}}x \right)-2x+2\sqrt{2}{{\tan }^{-1}}\left( \sqrt{2}\tan x \right)+c$

Note: Since this was a question of indefinite integral, the addition of a constant term is necessary. Do not forget to add the constant term at the end. Moreover, while solving such complex integrals you might miss out some terms or you might forget to integrate some terms, which would lead to an invalid expression. So, be very careful while solving for such integrals.