Intergration of Trig

Can someone please show how to do these type of questions?

Intergrate cos^4(x)sin^3(x) dx

Well with that one you use sin^2x = 1 - cos^2x

Then you have 2 parts that look like cos^nxsinx so you use inverse chain rule

cos^4(x)sinx(sin^2(x))
cos^4(x)sinx(1-cos(x))]

By the way its not cos to the power of 4x its cos to the power of 4

Yeah I know

Your working above is incorrect sin^2x=1-cos^x

Then multiply the bracket

cos^4(x)sinx(1-cos^2x)
cos^4(x)sinx -cos^6(x)sinx

Its this ok so far, what the next step?

think ive got it

i did

cos^4(x)sinx(1-cos^2(x)) then u = cos x u' = -sinx dx/du = -cosec x

put u in

- u^4 + u^6 then it grate and substitute cos x back in after

Will the same technique apply to intergrating sin^3x dx ?

my steps so far

Sinx(sin^2x)
Sinx(1-cos^2x) Now I cant see what to do?

Expand the bracket. Now integrate term by term (you can use a substitution for the second term).

I think im getting the hang of it, but it always seems to get harder could you show how you would solve this please

Intergrate in(rt(x-1))dx

Why does everyone think the natural logarithm is written as "in"?

Make a substitution $u=x-1$

Remember than $\ln(x^n)=n \ln x$

Integrate by parts.

Haha I dunno why everyone does, It just one of those things i guess

One more question I'm attempting to intergrate x^5e^(x^3)

I intergrate by parts but it just goes round in circle (x keeps coming back) is there a special technique for these type of questions?

You just need lots of experience.

Write $x^5$ as $x^3 x^2$ and make a substitution $u=x^3$.

so is dv/dx = x^2e^(x^3)?

v = 1/3e^(x^3)?

so is dv/dx = x^2e^(x^3)?

v = 1/3e^(x^3)?

You seem to have failed to make the substitution I suggested?

the workings looks horrible, maybe theres a quicker method?

Maybe MAKING A SUBSTITUTION would help?



$\displaystyle \int x^5 e^{x^3} \, dx = \frac{1}{3} \int t e^t \, dt$

I did substitute u = x^3 in, If you didnt mean it like this i dont understand im a novice

Maybe MAKING A SUBSTITUTION would help?



$\displaystyle \int x^5 e^{x^3} \, dx = \frac{1}{3} \int t e^t \, dt$

To be fair to Dopey, he saw your breakdown of x^5 with u=x^3 and thought you were doing IBP with dv/dx equal the rest.