Question

1. derive the reduction formulas for
2. ∫cosⁿxdx satisfies nI_n=cos^n-1xsinx+(n-1)I_n-2
3. ∫xⁿe^xdx
4. ∫tanⁿxdx

Answer 1

Solution

2) $∫cos^nxdx$

We have

$I_n=∫cos^nxdx$

$cos^n x=cos^{n-1}xcosx$

$\implies I_n=∫cos^{n-1}xcosxdx$

Let's do integration by parts

Let $u=cos^{n-1}x \implies du=(n-1)cos^{n-2}x(-sinx)dx=-(n-1)cos^{n-2}x(sinx)dx$

Let $dv=cosxdx \implies v=sinx$

$I_n=\int udv=uv-\int vdu$

$=cos^{n-1}xsinx-\int sinx(-(n-1)cos^{n-2}x(sinx))dx$

$=cos^{n-1}xsinx+\int sinx(n-1)cos^{n-2}xsinxdx$

$=cos^{n-1}xsinx+\int (n-1)cos^{n-2}xsinxsinxdx$

but $sinxsinx=sin^2x$

$\implies I_n=cos^{n-1}xsinx+\int (n-1)cos^{n-2}xsin^2xdx$

We know that $sin^2x=1-cos^2x$

$\implies I_n=cos^{n-1}xsinx+\int (n-1)cos^{n-2}x(1-cos^2x)dx$

$=cos^{n-1}xsinx+(n-1)\int cos^{n-2}x(1-cos^2x)dx$

$=cos^{n-1}xsinx+(n-1)\int cos^{n-2}xdx-(n-1)\int cos^nxdx$

$=cos^{n-1}xsinx+(n-1)I_{n-2}-(n-1)I_n$

Now, collect the like terms

$\implies I_n+(n-1)I_n=cos^{n-1}xsinx+(n-1)I_{n-2}$

$\implies I_n+nI_n-I_n=cos^{n-1}xsinx+(n-1)I_{n-2}$

$\therefore nI_n=cos^{n-1}xsinx+(n-1)I_{n-2}$

$\implies I_n={cos^{n-1}xsinx+(n-1)I_{n-2}\over n}$ for $n\ge 2$

For initial values $(I_2,I_3....)$

3) $∫x^ne^xdx$

Let's do integration by parts

Let $u=x^n \implies du=nx^{(n-1)}dx$

$dv=e^xdx \implies v=e^x$

$I_n=\int udv=uv-\int vdu$

$=x^ne^x-\int e^xnx^{(n-1)}dx$

$=x^ne^x-n\int x^{(n-1)}e^xdx$

$=x^ne^x-nI_{n-1}$

$\therefore I_n=x^ne^x-nI_{n-1}$ for $n\ge1$

For initial values $(I_1,I_2...)$

4) $∫tan^nxdx$

We have

$I_n=∫tan^nxdx$

$=∫tan^{n-2}xtan^2xdx$ but $tan^2x=sec^2x-1$

$=∫tan^{n-2}x(sec^2x-1)dx$

$=∫tan^{n-2}xsec^2xdx-\int tan^{n-2}xdx$

Let's evaluate $∫tan^{n-2}xsec^2xdx$

Let $u=tanx \implies du=sec^2xdx$

Rewrite $∫tan^{n-2}xsec^2xdx$ in terms of $u$ as follows

$∫tan^{n-2}xsec^2xdx=\int u^{n-2}du={u^{n-1}\over n-1}={tan^{n-1}x\over n-1}$

$\implies I_n={tan^{n-1}x\over n-1}-\int tan^{n-2}xdx$

$\therefore I_n={tan^{n-1}x\over n-1}-I_{n-2}$ for $n\ge2$

For initial values $(I_2,I_3...)$