Question

Integration by parts

1. Integral of 5 sin^-1 (x) dx

2. Integral of 2 lnx dx

Answer 1

Calculus

We need to find the Integration by parts.

Solution:

We know the formula of Integration by Parts

$\int f(x) \space g(x) \space dx = f(x) \space \int g(x) dx - \int f'(x) (\int g(x) dx) \space dx$

1).

$\int 5 sin^{-1} x dx$ = 5 $\int sin ^{-1} x \times 1 \space dx$

Here, f(x) = $sin^{-1} x$ and g(x) = 1 ( By Using ILATE)

$\int 5 sin^{-1} x dx$ = 5 [ $sin^{-1} x \int 1 dx - \int d(sin^{-1} x) (\int 1 dx) dx$ ]

$$= 5 [ x sin^{-1} x - \int \frac {1}{\sqrt {1-x^2}}\times x dx ]$$

$$= 5 [x sin^{-1} x - \frac {1}{2} \int \frac {2x}{\sqrt {1-x^2}} dx]$$

$$= 5 [ x sin^{-1} x + \frac {1}{2} \int t^ {\frac {-1}{2}} dt] + c$$

Here $t = 1 - x^2 \\ dt = - 2x dx$

$\int 5 sin^{-1} x dx$ =

$$5 [ x sin^{-1} x + \frac {1}{2} \frac {t^{\frac {-1}{2} + 1}}{\frac {-1}{2}+1}] + c$$

$$= 5 [ x sin^{-1} x + t^{ \frac {1}{2}}] + c = 5 [ x sin^{-1} x + \sqrt {1-x^2}] +c$$

2).

$\int 2 ln x = 2 \int ln x \times 1 \space dx$

Using ILATE , let f(x) = ln x and g(x) = 1

$$\int 2 ln x = 2 [ln x \int 1 dx- \int f' (x) (\int g(x) dx )dx]$$

$$= 2 [ x \space ln x - \int \frac {1}{x} \times x dx ] + c$$

$$= 2 [ x \space ln x - x ] + c$$