Methods of Integration> Integration by Parts

If two functions are written in the form uu and dvdv, then integration by parts is based on the product rule of differentiation.

\[\int\left(\mathrm{u.v}\right)\mathrm{dx}=\mathrm{u}\int\mathrm{v}\mathrm{dx}-\int\left(\frac{\mathrm{du}}{\mathrm{dx}}\right).\left(\int\mathrm{v}\mathrm{dx}\right)\mathrm{dx}\]

Find \[\int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} dx\]

Solution: Let first function be \[\sin^{-1} x\] and second function be \[\frac{x}{\sqrt{1 - x^2}}\].

First, we find the integral of the second function, i.e., \[\int \frac{x dx}{\sqrt{1 - x^2}}\].

Use substitution.

Put \[t = 1 - x^2\]. Then \[dt = -2x dx\]

Therefore,

Hence,

Apply Integration by Parts

Using

Alternatively, this integral can also be worked out by making the substitution \[\sin^{-1} x = \theta\] and then integrating by parts.

Find \[\int \sqrt{3 - 2x - x^2} dx\]

Solution: Note that \[\int \sqrt{3 - 2x - x^2} dx = \int \sqrt{4 - (x + 1)^2} dx\]

Put \[x + 1 = y\] so that \]dx = dy\].

Thus

• Formula:

  \[\int u dv = uv - \int v du\]

• Choose u by LIATE

• For log x and inverse trig, multiply by 1

• Repeated parts may be needed for \[e^x \sin x\], \[e^x \cos x\].