Question

I don't know how to write integral sign so I substitute it with S

Use integration by part to prove the formula
S(cos^nx)dx = 1/n(cos^(n-1)xsinx + (n-1)/n S(cos^(n-2)x)dx

Answer 1

can you write your question clearly

Answer 2

It's because i don't know how to write sign of integration, so I use S as integral sign

Answer 3

\[\int\limits_{}^{}\cos(x)^n.dx = \frac{1}{n}(\cos^{n-1}x.sinx + \frac{n-1}{n}\int\limits_{}^{}\cos^{n-2}x).dx\]

Is it???

Answer 4

\[\int\limits(\cos^nx)dx = 1/n(\cos^{n-1}x)(\sin{x}) + (n-1)/n \int\limits(\cos^{n-2}x)dx\]

Answer 5

See,

\[\cos^nx = x^0.\cos^nx\]

Now use ILATE rule here..
x^0 is the first function and cos^nx is the second function..
Use the formula for Integration by parts and proceed further...

Answer 6

\[\int\limits_{}^{}\cos^n(x)dx = \int\limits_{}^{}\cos^{n-1}(x)\cos(x)dx\]
Let cos(x) be u' and cos^(n-1)(x) be v
\[\int\limits_{}^{}\cos^{n-1}(x)\cos(x)dx = \cos^{n-1}(x)\sin(x) - \int\limits_{}^{}-(n-1)\cos^{n-2}(x)\sin(x)\sin(x)dx = \]
\[\cos^{n-1}(x)\sin(x) + (n-1)\int\limits_{}^{}\cos^{n-2}(x)\sin^2(x)dx\]

Use the identity \[\sin^2(x) = 1-\cos^2(x)\]
\[\int\limits_{}^{}\cos^{n}(x)dx = \cos^{n-1}(x)\sin(x) + (n-1)\int\limits_{}^{}\cos^{n-2}(x)(1-\cos^2(x))dx =\cos^{n-1}(x)\sin(x) + (n-1)\int\limits_{}^{}\cos^{n-2}(x)-\cos^n(x)dx \]
\[\cos^{n-1}(x)\sin(x) + (n-1)\int\limits_{}^{}\cos^{n-2}(x)-\cos^n(x)dx \]
\[\int\limits_{}^{}\cos^n(x)dx = \cos^{n-1}(x)\sin(x) + (n-1)\int\limits_{}^{}\cos^{n-2}(x)-\cos^n(x)dx\]\[\int\limits_{}^{}\cos^n(x)dx = \cos^{n-1}(x)\sin(x) + (n-1)\int\limits_{}^{}\cos^{n-2}(x) - (n-1) \int\limits_{}^{}\cos^n(x)dx\]
\[(1 + n -1) \int\limits_{}^{}\cos^n(x)dx = \cos^{n-1}(x)\sin(x) - (n-1)\int\limits_{}^{}\cos^{n-2}(x)\]
\[ \int\limits\limits_{}^{}\cos^n(x)dx = { \cos^{n-1}(x)\sin(x) \over n } + {(n-1)\over n}\int\limits\limits_{}^{}\cos^{n-2}(x)\]

Answer 7

\[\int\limits_{}^{}uv.dx = u \int\limits_{}^{}v.dx - \int\limits_{}^{}(\frac{du}{dx} \times \int\limits_{}^{}v.dx).dx\]

Answer 8

Woops. Some of it got cut off. I hope you can still make sense out of it. Big equations :(

Answer 9

on the equation before the last i put "- (n-1) intergral" etc, it should be a + sign before (n+1), but i think you understand that too :)

the equation thing gets messy hehe.

Answer 10

ahah what is wrong with me today? a + sign before (n MINUS 1)

Answer 11

Thanks a bunch @soati and @waterineyes

Answer 12

Welcome dear...