How do I find this integral, in as much detail as possible please?
$$\int x \cos (x^2) \,\mathrm dx $$
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Use $u=x^2$ and then use integration by parts. I believe you can do the rest :). – Chinny84 Apr 30 '14 at 11:57
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2@Chinny84 , after putting $u=x^2$, there will be no need for integration by parts – idpd15 Apr 30 '14 at 11:59
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@MridulSachdeva oops! – Chinny84 Apr 30 '14 at 12:06
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@leiahassan it is good practice to give some idea of what you have already tried when asking a question – Joe Tait Apr 30 '14 at 12:13
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@JoeTait hehe sorry, this is my first time using stack exchange. Thanks for your help everyone. Got it now. – Leia Hassan Apr 30 '14 at 12:20
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If we put $$u = x^2 $$ then $$ du = 2x\,dx \iff xdx = \frac 12 du$$
Then our integral becomes $$\int \cos \underbrace{(x^2)}_u \,\underbrace{x\;dx}_{\frac 12\; du} = \frac 12 \int \cos(u)\,du = \frac 12 \sin\underbrace{(u)}_{x^2} + C$$
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Let $t=x^2$ so $dt=2xdx$ hence $$\int x\cos\left( x^2\right)dx=\frac12\int \cos(t)dt=\frac12\sin (t)+C=\frac12\sin\left(x^2\right)+C$$