Answer to Question #103231 in Calculus for Patrick Kariuki

Question

Integrate tan³x dx using reduction method

Answer 1

$I=\int \:tan^3\left(x\right)dx$

$=\int \:tan^2\left(x\right)\cdot tan(x)\:dx$

We know $tan^2(x)=sec^2(x)-1$ . Substituting this in above integral.

$=\int \:\left(sec^2x-1\right)tan\left(x\right)dx$

$=\int \:\left(\:tan\left(x\right)sec^2\left(x\right)-\:tan\left(x\right)\right)dx$

$=\int \:tan\left(x\right)sec^2\left(x\right)dx-\int \:tan\left(x\right)dx$

$=I_1-I_2$ ......................................................................................................eqn (1)

$I_1=\int tan(x)sec^2(x)dx$

apply u-substitution

Let $u=tanx$

$\Rightarrow du=sec^2(x)dx$

$I_1=\int u du$

$=\frac{u^2}{2}+C_1$ where $C_1$ = integral constant

substituting $u = tanx$ back

$I_1=\frac{tan^2(x)}{2}+C$

Also $I_2=\int tan(x)dx=-ln|cos(x)|+C_2$ where $C_2$ = integral constant

Substituting value of $I_1$ and $I_2$ in equation 1

$I=\frac{tan^2(x)}{2}+ln|cos(x)|+C$ where $C=C_1+C_2,$ another integration constant.

Another method: Using direct reduction formula (if you know or remember, otherwise please use above method).

$\int \:tan^n\left(x\right)dx=\frac{tan^{n-1}\left(x\right)}{n-1}-\int \:tan^{n-2}\left(x\right)dx+C$ where $n\ne1$

Putting n =3

I $=\frac{tan^2\left(x\right)}{2}-\int \:tan\left(x\right)dx$

I $=\frac{tan^2\left(x\right)}{2}-(-ln|cos(x)|)+C$

I $=\frac{tan^2(x)}{2}+ln|cos(x)|+C$