MHB What is the integral of sin^3(t)cos^4(t) using u-substitution?

[karush]

Whit 8.7.23} trig u subs s87.3 nmh{1000}
$\displaystyle
I=\displaystyle\int {\sin^3\left({t}\right) \cos^4\left({t}\right)} \ d{t}
=\int\ (1-\cos^2\left({t}\right)) \cos^4\left({t}\right) \sin\left({t}\right) \ dt \\
\begin{align}\displaystyle
u& = \cos\left({t}\right)&
du&=-\sin\left({t}\right) \ d{t} \\
\end{align}\\

I=-\displaystyle\int\left(1-u^2\right)u^4 \ du = - \displaystyle\int\left(u^4-u^6\right) \ du \\
\text{integrate }\\
I =-\left[ { \dfrac{u^5}{5}}
-\dfrac{u^7}{7}\right] + C \\
\text{back substitute }\\
I = { \dfrac{\cos^7{t} }{7}}
-\dfrac{\cos^5\left({t}\right)}{5} + C$
Hopefully

 

[Greg]

That's it. Good work!

 

[karush]

Much Mahalo

MHB has done a lot to help me prepare for the MAT 206 class coming up in August

Just curious how can an answer be checked without using a calculator which you can't do during a test.

 

[Greg]

Differentiate the result.

 

[Prove It]

Was this the entire question? Or was it an integral resulting from having done a trigonometric substitution?

 

[karush]

Both