Methods of Integration> Integration by Parts

Integration by parts is a method of integration based on the product rule of differentiation.

If u and v are differentiable functions, then \[ \boxed{\int u\, dv = uv - \int v\, du} \]

For a product f(x)g(x), \[ \boxed{\int f(x)g(x)\, dx = f(x)\int g(x)\, dx - \int\left[f'(x)\int g(x)\, dx\right]dx} \]

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\[ \int x \cos x\, dx. \]

Solution:

Take x as the first function and cos x as the second function.

Using integration by parts,

\[ \int x \cos x\, dx = x \int \cos x\, dx - \int \left(\frac{d}{dx}x\right)\left(\int \cos x\, dx\right) dx \]

\[ = x \sin x - \int \sin x\, dx \]

Therefore, \[ \int x \cos x\, dx = x \sin x + \cos x + C \]

Find \[ \int e^x \sin x\, dx. \]

Solution:

Let \[ I = \int e^x \sin x\, dx. \]

Applying integration by parts,

\[ I = -e^x \cos x + \int e^x \cos x\, dx. \]

Again applying integration by parts,

\[ \int e^x \cos x\, dx = e^x \sin x - I. \]

Thus,

\[ I = -e^x \cos x + e^x \sin x - I \]

\[ 2I = e^x(\sin x - \cos x). \]

Hence, \[ \int e^x \sin x\, dx = \frac{e^x}{2}(\sin x - \cos x) + C \]

Integral of the Form ∫ eˣ[f(x) + f'(x)] dx

If an integral is of the form

\[ \int e^x[f(x) + f'(x)]\, dx, \]

then

\[ \boxed{\int e^x[f(x) + f'(x)]\, dx = e^x f(x) + C} \]

because

\[ \frac{d}{dx}[e^x f(x)] = e^x f(x) + e^x f'(x) = e^x[f(x) + f'(x)]. \]

• 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 \[e^x \sin x\], \[e^x \cos x\].