integrate $\int \frac{\tan^4x}{4}\cos^3x$ 
$$\int \frac{\tan^4x}{4}\cos^3x$$

$$\int \frac{\tan^4x}{4}\cos^3x=\frac{1}{4}\int \frac{\sin^4x}{\cos^4x}\cos^3x=\frac{1}{4}\int\frac{\sin^4x}{\cos x}=\frac{1}{4}\int\frac{\sin^2x\cdot\sin^2x}{\cos x}=\frac{1}{4}\int\frac{\sin^2x\cdot(1-\cos^2x)}{\cos x}=\frac{1}{4}\int \frac{\sin^2x}{\cos x}-\frac{\cos^2x}{\cos x}=\frac{1}{4}\int \frac{1-\cos^2x}{\cos x}-\cos x=\frac{1}{4} \int \frac{1}{\cos x}-2\cos x=\frac{1}{4}(\ln(\tan x+\sec x)+2\sin x$$
Is it correct?
 A: No, you've made a mistake in your fifth step! Solve it like this:
$$\int\frac{\tan^4(x)\cos^3(x)}{4}\space\text{d}x=\frac{1}{4}\int\tan^4(x)\cos^3(x)\space\text{d}x=\frac{1}{4}\int\tan(x)\sin^3(x)\space\text{d}x=$$

Substitute $u=\sin(x)$ and $\text{d}u=\cos(x)\space\text{d}x$:

$$\frac{1}{4}\int-\frac{u^4}{u^2-1}\space\text{d}u=-\frac{1}{4}\int\left[1+u^2+\frac{1}{2(u-1)}-\frac{1}{2(u+1)}\right]\space\text{d}u=$$
$$-\frac{1}{4}\left[\int1\space\text{d}u+\int u^2\space\text{d}u+\frac{1}{2}\int\frac{1}{u-1}\space\text{d}u-\frac{1}{2}\int\frac{1}{u+1}\space\text{d}u\right]=$$

Notice:


*

*$$\int 1\space\text{d}y=y+\text{C}$$

*$$\int y^n\space\text{d}y=\frac{y^{n+1}}{n+1}+\text{C}$$


$$-\frac{1}{4}\left[u+\frac{u^3}{3}+\frac{1}{2}\int\frac{1}{u-1}\space\text{d}u-\frac{1}{2}\int\frac{1}{u+1}\space\text{d}u\right]=$$



*

*Substitute $p=u-1$ and $\text{d}p=\text{d}u$

*Substitute $s=u+1$ and $\text{d}s=\text{d}u$


$$-\frac{1}{4}\left[u+\frac{u^3}{3}+\frac{1}{2}\int\frac{1}{p}\space\text{d}p-\frac{1}{2}\int\frac{1}{s}\space\text{d}s\right]=-\frac{1}{4}\left[u+\frac{u^3}{3}+\frac{\ln\left|p\right|}{2}-\frac{\ln\left|s\right|}{2}\right]+\text{C}=$$
$$-\frac{1}{4}\left[u+\frac{u^3}{3}+\frac{\ln\left|u-1\right|}{2}-\frac{\ln\left|u+1\right|}{2}\right]+\text{C}=$$
$$-\frac{1}{4}\left[\sin(x)+\frac{\sin^3(x)}{3}+\frac{\ln\left|\sin(x)-1\right|}{2}-\frac{\ln\left|\sin(x)+1\right|}{2}\right]+\text{C}=$$
$$-\frac{1}{4}\left[\frac{3\sin(x)+\sin^3(x)}{3}+\frac{\ln\left|\sin(x)-1\right|-\ln\left|\sin(x)+1\right|}{2}\right]+\text{C}$$
A: \begin{align}
  \frac{1}{4}\int \tan^4 x \cos^3 x \, dx
    &= \frac{1}{4} \int \frac{\sin^4 x}{\cos^4 x}\cos^3 x \, dx\\[0.3cm]
    &= \frac{1}{4} \int \frac{\sin^4 x}{\cos^2 x}\cos x \, dx\\[0.3cm]
    &= \frac{1}{4} \int \frac{\sin^4 x}{1-\sin^2x} \cos x \, dx
\end{align}
Now do a $u$-sub with $u = \sin x$.
A: Omitting the $\frac{1}{4}$, we have that
$\displaystyle\tan^4 x\cos^3 x=\frac{\sin^4 x}{\cos x}=\frac{(1-\cos^2 x)^2}{\cos x}=\frac{1-2\cos^2 x+\cos^4 x}{\cos x}=\sec x-2\cos x+\cos^3 x$
$\hspace{.85 in}\displaystyle=\sec x-2 \cos x+\cos x(1-\sin^2x)=\sec x-\cos x-\sin^2 x\cos x$, 
so $\displaystyle\int\tan^4x\cos^3x \;dx=\ln|\sec x+\tan x|-\sin x-\frac{1}{3}\sin^3x+C$
A: More easy way $tan(x).cos(x)=sin(x)$ so integral becomes $sin^3(x).tan(x)=sin^4(x)/cos(x)$ Now $sinx=u$ so integral becomes $$\int\frac{u^4}{1-u^2}$$ which on simplification becomes $$\int \frac{2}{(1/u^2)(1/u^2-1)}du$$  now let $1/u=t$ thus the common $ u^2$ cancels to give $\int\frac{1}{1-t^2}dt$ whose integral is $\frac{1}{2}log(\frac{1+t}{1-t})+c$ and the resubstitution gives the answer .
