## Integration by Parts

In this page integration by parts we are going to see where we have to apply this method. Integration-by-parts is one of the method basically used o find the integral when the integrand is a product of two different kind of function.

Formula :

∫ u dv  = uv - ∫ v du

∫ log x dx

∫ tan⁻ ¹ x dx

∫ xⁿ log x dx

∫ xⁿ tan⁻ ¹ x dx

∫ xⁿ e^(ax) dx

∫ xⁿ sin x dx

∫ xⁿ Cos x dx

log x

tan⁻ ¹ x

log x

tan⁻ ¹ x

xⁿ

xⁿ

xⁿ

# dv

dx

dx

xⁿ dx

xⁿ dx

e^(ax) dx

sin x dx

Cos x dx

Now let us see some example problems to understand this topic better.

Example 1:

Integrate x sec² x with respect to x

Solution:

In this problem we must apply the concept integrating-by-parts. Because here we have two function which are multiplying. To integrate this we must apply this concept.

∫ x sec² x dx

u = x                dv = sec² x dx

du = dx          ∫ dv = ∫ sec² x dx

v = tan x

Formula :

∫ u dv  = uv - ∫ v du

∫ x sec² x dx = x (tan x) - ∫ (tan x) dx

= x (tan x) - sec² x + C

Example 2:

Integrate Sin⁻¹ x with respect to x

Solution:

In this problem we must apply the concept integrating-by-parts. Because here we have two function which are multiplying. To integrate this we must apply this concept.

∫ Sin⁻¹ x dx

u = Sin⁻¹ x                    dv =  dx

du = 1/√(1-x²)          ∫ dv = ∫ dx

v =

Formula :

∫ u dv  = uv - ∫ v du

∫ Sin⁻¹ x dx = Sin⁻¹ x (x) - ∫ (x) [1/ √(1-x²)] dx

= x  Sin⁻¹ x - ∫ (x) [1/ √(1-x²)] dx -----(1)

Now we have to apply substitution method

t =  √(1-x²)

= (1-x²)

2t dt = (0 - 2x) dx

2t dt = -2x dx

t dt = -x dx

-t dt = x dx

= x  Sin⁻¹ x - ∫ [x dx/ √(1-x²)]

= x  Sin⁻¹ x - ∫ [-t dt/ t]

= x  Sin⁻¹ x + ∫ dt

= x  Sin⁻¹ x + t + C

= x  Sin⁻¹ x + √(1-x²) + C  integration by parts integration by parts

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