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sha0403
help me to integrate this.. integration of cos^3 x dx..
split the cos^3x into cosx and cos^2x then use the trigonometric substitution. Replace cos^2x by (1 - sin^2x). Now you can use some integrate it.
but i still did not get sin x - 1/3 sin^3 (x)+c... bcoz the question ask to prove..
Look. When you did the substitutions as I told you above, you should end up with the following integral:\[\int\limits_{}^{} [(1 - \sin^{2}(x) )\cos(x) ]dx\]
int (1-sin^2 x)cosx dx = int cosx dx - int sin^2 x cosx dx for int cosx dx = .... and to find int sin^2 x cosx dx, use int by sub, let u=sinx du = .... continue it
ok the i get integration of (cos x - u^2) du.. how to solve integration of cos x in term of u?
if u = sin(x), then du = cos(x)dx. So replace the sin^2(x) in the original integral by u^2 and cos(x)dx by du. You will have a really trivial integral.
\[\int\limits(\cos ^{3}{x})=\int\limits(cosx*\cos^{2}{x})=\int\limits(cosx(1-\sin ^{2}{x})=\int\limits(cosx)-(sinx)^2(cosx)=sinx-1/3(\sin ^{3}{x})+\]
a proof for my answer: cos^3x can be written as cos^2x*cosx. Then, we all know that cos^2x can be written as 1-sin^2x. then we solve the bracket. then we'll find (sinx)^2*cosx , this remind us of the integration rule where [f(x)]^n*f'(x) = ([f(x)]^n+1/n+1)+c
but how to solve the integration of cos x-u^2 du?
no.. like this integration of (cos x)-(u^2) du
@sha0403\[\Large \int\limits (1- \sin^2 x)(\cos x dx)\]let \(u=\sin x\), \(du = \cos x dx\)
yes but how ti integrate cos x in term of du?
use the substitution method.
ok i get it now.. thank u..sorry i was too slow..
\[\cos^2x*cosx dx=\cos^3x \]