Integration By Parts Formula - Practice Questions & MCQ

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Quick Facts

Integration by Parts is considered one the most difficult concept.
43 Questions around this concept.

Solve by difficulty

MEDIUM

If $\int f(x)dx=\Psi (x),then\; \int x^{5}f(x^{3})dx$ is equal to :

$\frac{1}{3}\left [ x^{3}\Psi \left ( x^{3} \right )-\int x^{3}\Psi (x^{3})dx \right ]+C$

$\frac{1}{3}\left [ x^{3}\Psi \left ( x^{3} \right )-\int x^{2}\Psi (x^{3})dx \right ]+C$

$\frac{1}{3}\, \, x^{3}\Psi \left ( x^{3} \right )-3\int x^{3}\Psi (x^{3})dx +C$

$\frac{1}{3}\, \, x^{3}\Psi \left ( x^{3} \right )-\int x^{2}\Psi (x^{3})dx +C$

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If $2\int_{0}^{1}\tan ^{-1}xdx = \int_{0}^{1}\cot ^{-1}(1-x+x^{2})dx , then \int_{0}^{1}\tan ^{-1}(1-x+x^{2})dx$ is equal to :

log4

$\frac{\pi }{2}+log 2$

log2

$\frac{\pi }{2}-log 4$

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Concepts Covered - 2

Integration by Parts

To evaluate integration of a function which is a product of two functions, we use the method of integration by parts.

If two functions of x, u and v are given, then

$\mathrm{\int\mathit{uv\;dx=u\times\int v\;dx-\int\left \{ \frac{du}{dx}\int v\;dx \right \}dx}}$

i.e.

The integral of product of two functions =

( first function) X ( integral of second function ) - integral of (differential of first function X integral of second function).

Proof:

Let, h(x) = f(x) . g(x), then by using the product rule, we obtain, h′(x) = f ′(x) g(x) + g′(x) f(x). Let’s now integrate both sides of this equation:

$\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\int h^{\prime}(x) d x=\int\left(g(x) f^{\prime}(x)+f(x) g^{\prime}(x)\right) d x}\\\\\text{this gives us, }\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;h(x)=f(x) g(x)=\int g(x) f^{\prime}(x) d x+\int f(x) g^{\prime}(x) d x}\\\\\mathrm{\text { Now we solve for } \int f(x) g^{\prime}(x) d x}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\int f(x) g^{\prime}(x) d x=f(x) g(x)-\int g(x) f^{\prime}(x) d x}\\\\\text{By making the substitutions u = f(x) and v = g'(x) which in turn make }\\ g(x)=\int vdx \text{, we have the more compact form}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\int uv\; d x=u \int v\;dx-\int \left \{ \frac{du}{dx} \int v\;dx \right \}dx}$

Important point for selecting first function (u) and second function (v)

Usually we use the following preference order for selecting the first function. (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential). In the above stated order, the function on the left is always chosen as the first function. This rule is called as ILATE.

For example, if an integral contains a logarithmic function and an algebraic function, we should choose logarithmic function as the first function (u).

Application of of Integration by Parts

One of the interesting application of integration by part is:

$\mathbf{\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+C}$

Proof:

$\\\mathrm{We \;have\;given\;\;\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x}\\\\\mathrm{\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\int e^{x} \cdot f(x) d x+\int e^{x} \cdot f^{\prime}(x) dx}\\\\\mathrm{In\;first\;integral,\;take\;f(x)\;as\;first\;function\;and\;e^x\;as\;second\;function}\\\\\mathrm{\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=f(x) \cdot e^{x}-\int f^{\prime}(x) \cdot e^{x} d x+\int e^{x} \cdot f^{\prime}(x) d x+C}\\\\\mathrm{\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=f(x) \cdot e^{x}+C}$

$\\ {\text {Thus, to evaluate the integrals of the type }}{ \int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x}\\ \text {we first express the integral as the sum of two integrals }\int e^{x} f(x) d x\\ \text { and } \int e^{x} f^{\prime}(x) d x \text { and then integrate the }\text { integral involving } e^{x} f(x) \\\text { as integral by parts taking } e^{x}\text{ as second function.}$

General Formula

$\mathbf{\int e^{g(x)} \cdot\left\{f(x) g^{\prime}(x)+f^{\prime}(x)\right\} d x=f(x) \cdot e^{g(x)}+C}$

$\mathbf{Integration\; of \;\int e^{ax}\cdot \sin (bx)\;dx\;\;and\;\;\int e^{ax}\cdot \cos (bx)\;dx}$

$\\\mathrm{Let,\;\;\;\;\;\;\;\;\;\;I=\int e^{a x}(\sin b x) d x}\\\text{take }\sin bx \;\mathrm{as\;first\;function\;and\;e^{ax}\;as\;second}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\sin b x \cdot\left(\frac{e^{a x}}{a}\right)-\int b \cos b x \cdot \frac{e^{a x}}{a} d x}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{1}{a} \sin b x \cdot e^{a x}-\frac{b}{a}\left\{\cos b x \cdot \frac{e^{a x}}{a}-\int(-b \sin b x) \cdot \frac{e^{a x}}{a} d x\right\}}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{1}{a} \sin b x \cdot e^{a x}-\frac{b}{a^{2}} \cos b x \cdot e^{a x}-\frac{b^{2}}{a^{2}} \int \sin b x \cdot e^{a x} d x}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;I=\frac{1}{a} \sin b x \cdot e^{a x}-\frac{b}{a^{2}} \cos b x \cdot e^{a x}-\frac{b^{2}}{a^{2}} I}\\\mathrm{\therefore \quad I+\frac{b^{2}}{a^{2}} I=\frac{1 \cdot e^{a x}}{a^{2}} \cdot(a \sin b x-b \cos b x)}\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;I=\frac{e^{a x}}{a^{2}+b^{2}}(a \sin b x-b \cos b x)+C}$