Techniques of integration

[hrr379]

 

[tkhunny]

u ==> du is a derivative, not an antiderivative.

\(\displaystyle u = \sqrt{x}\)

\(\displaystyle du = \frac{1}{2\sqrt{x}}\)

Try to keep track of which direction you are going.

 

[hrr379]

Still lost, sir.

 

[Deleted member 4993]

View attachment 2017

u = √x

du = 1/(2√x) dx → dx/√x = 2 du

\(\displaystyle \int\dfrac{sin^3(\sqrt{x})}{\sqrt{x}} dx \ = \ 2 * \int sin^3(u) du \) ................. continue.....

 

[hrr379]

u = √x

du = 1/(2√x) dx → dx/√x = 2 du

\(\displaystyle \int\dfrac{sin^3(\sqrt{x})}{\sqrt{x}} dx \ = \ 2 * \int sin^3(u) du \) ................. continue.....

Thank you.