Evaluate $\int \cos^2(x)\tan^3(x) dx$ using trigonometric substitution How would I integrate to evaluate $\int \cos^2(x)\tan^3(x) dx$ using trigonometric substitution?
I made an attempt by making substitutions such as $$\cos^2(x)=1-\sin^2(x)$$
$$\tan(x) = \frac{\sin(x)}{\cos(x)}$$ and $$\tan^2(x)=1+\sec^2(x)$$
But I couldn't find a way to make it look like an integral I could solve using a $u$ substitution or identity.
Could I get some help on this one?
 A: Use the definition of $\tan x$, then after a bit of algebra you have 
$$
\frac{\sin x }{\cos x} - \cos x \sin x
$$
The first one is solved using logarithm subsitution, the second using the identity $2 \sin x \cos x = \sin 2 x$
A: $$\int \left (\cos^{2}x  \right )\cdot \left (\tan^{3}x  \right )dx$$$$ =
\int \left (\cos^{2}x  \right )\cdot \left (\frac{\sin^{3}x }{\cos^{3}x} \right)dx$$
$$=\int \left (\frac{\sin^{3}x }{\cos x} \right )dx$$$$
= \int \left (\frac{\left (\sin^{2}x   \right )\sin x}{cos x} \right )dx$$
$$=
\int \left (\frac{\left (1- \cos^{2}x   \right )\sin x}{\cos x} \right )dx$$$$=
\int \left [\left (\frac{\sin x}{\cos x} \right )-\left ( \cos x\cdot \sin x \right )  \right ]dx= ...$$
A: Simplify: $$\cos^2x\tan^3x=\cos^2x\frac{\sin^3x}{\cos^3x}=\frac{\sin^3x}{\cos x}$$
Assume that $$\cos x=u, \ \  \ dx=-\frac{du}{\sin x}$$
$$\int \frac{\sin^3xdx}{\cos x}=\int \frac{\sin^3xdx}{u}\frac{-du}{\sin x}$$
$$=-\int \frac{(1-\cos^2x)}{u} du$$
$$=-\int \frac{(1-u^2)}{u} du$$
$$=\int (u-\frac{1}{u}) du$$
$$=\frac{u^2}{2}-\ln|u|+C$$
$$=\frac{\cos^2x}{2}-\ln|\cos x|+C$$
Where, $C$ is a constant for integration.
A: $$\int \cos^2x\tan^3x\ dx=\int \frac{\sin^3x}{\cos x}\ dx=\int \frac{(1-\cos^2x)\sin x}{\cos x}\ dx$$
Let $\cos x=t\implies -\sin x\ dx=dt$
$$=\int \frac{(t^2-1)dt}{t}$$
$$=\int \left(t-\frac{1}{t}\right)dt$$
A: Just for fun:
$$\int \cos^2 x \tan^3x  \ \mathrm{d} x$$
$$=\int \frac{\tan^3 x}{\sec^2 x} \ \mathrm{d} x$$
Now let $u = \tan^2 x, \mathrm du = 2 \tan x \sec^2 x \ \mathrm dx$:
$$=\int \frac{u \tan x }{\sec^2 x} \cdot \frac{\mathrm{d} u}{2 \tan x \sec^2 x}$$
$$=\frac{1}{2} \int \frac{u}{\sec^4 x} \mathrm d u$$
$$=\frac{1}{2} \int \frac{u}{(1+u)^2} \ \mathrm d u$$
$$=\frac{1}{2} \int \frac{1+u}{(1+u)^2} -\frac{1}{(1+u)^2} \mathrm d u$$
$$=\frac{1}{2} \left( \ln(\tan^2 x + 1) + \frac{1}{1+\tan^2 x} \right)  +C$$
$$=\frac{1}{2} \left( \ln | \sec^2 x| + \cos^2 x \right) + C$$
where we have used the identity $1 + \tan^2 x = \sec^2 x$ twice.
Further simplifying gives the accepted answer:
$$=\frac{1}{2} \left( \ln | \cos^{-2} x| + \cos^2 x \right) + C$$
$$=\frac{1}{2} \left(-2 \ln | \cos x | + \cos^2 x \right) + C$$
$$=\frac{\cos^2 x}{2} - \ln | \cos x | + C$$
