Integrate $\int \frac {\cos^2(x)}{1+\tan(x)} dx$

Question

I have to integrate this : $$\int \frac {\cos^2(x)}{1+\tan(x)} dx $$ My attempt: I assumed $\tan (x)=t$ and wrote $\cos^2(x)$ as $\frac 1{1+\tan^2(x)}$ With this I ended up with $$\int \frac {1}{(1+t) (1+t^2)^2} dt $$ How do I solve this integration in a simpler method or do I continue my method and just separate the denominator to simplify further?

Answer 1

We need to calculate $$ I=\int\frac{\cos^2x}{1+\tan x}\,dx=\int\frac{\cos^3x}{\sin x+\cos x}\,dx. $$ Consider another integral $$ J=\int\frac{\sin^3x}{\sin x+\cos x}\,dx $$ and calculate $I+J$ and $I-J$ using $$ a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2). $$ \begin{align} \color{blue}{I+J}&=\int(\cos^2x-\cos x\sin x+\sin^2x)\,dx=\int\left(1-\frac12\sin 2x\right)\,dx=\color{blue}{x+\frac{\cos 2x}{4}}+C,\\ \color{red}{I-J}&=\int\frac{\cos x-\sin x}{\sin x+\cos x}\cdot(\cos^2x+\cos x\sin x+\sin^2x)\,dx=\\ &=\int\frac{\cos x-\sin x}{\sin x+\cos x}\cdot\frac{(\cos x+\sin x)^2+1}{2}\,dx=\\ &=\frac12\int\cos 2x\,dx+\frac12\int\frac{\cos x-\sin x}{\sin x+\cos x}\,dx=\color{red}{\frac{\sin 2x}{4}+\frac12\ln|\sin x+\cos x|}+C. \end{align} Now add $\color{blue}{I+J}$ and $\color{red}{I-J}$ and divide by two.

Answer 2

for comparison your indefinite integral is given by $$\frac{1}{8} \left(\frac{2 (t+1)}{t^2+1}-\log \left(t^2+1\right)+2 \log (t+1)+4 \tan ^{-1}(t)\right)$$

Answer 3

HINT:

$\cos x+\sin x=\sqrt2\cos\left(x-\dfrac\pi4\right)$

Set $x-\dfrac\pi4=u$

Now $\cos x=\cos\left(\dfrac\pi4+u\right)=\dfrac{\cos u-\sin u}{\sqrt2}$

Now use $(a-b)^3$ expansion

Finally for $\dfrac{\sin^3 u}{\cos u}=\dfrac{\left(1-\cos^2u\right)\sin u}{\cos u}$ $=\tan u-\cos u\sin u$

and use $\sin2u=2\sin u\cos u$