Learn formula for integration by parts with examples. Use integration by parts formula UV & exponential functions to evaluate indefinite and definite Integrals. Practice problems in which Integration by Parts is to be done more than once.

- Integration by parts is a technique used for integration of the product of two functions.
- The integration by parts corresponds to the product rule of differentiation just like substitution rule in integration corresponds to chain rule in differentiation.

- If \(u(x)\) and \(v(x)\) are the two functions of \(x,\) then by the product rule of differentiation

\(\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\)

\(d(uv)=u\dfrac{dv}{dx}dx+v\dfrac{du}{dx}dx\)

\(d(uv)=udv+vdu\)

Now integrating both sides,

\(\int d(uv)=\int udv+\int vdu\)

\(uv=\int udv+\int vdu\)

\(\int udv=uv-\int vdu\)

This is the formula for integration by parts.

- Another form of formula for integration by parts for two functions \(f(x)\) and \(g(x)\) is given as

\(\int f(x)\,g'(x)\,dx=f(x)\,g(x)-\int f'(x)\,g(x)\,dx\)

A \(\dfrac {-x\;cos5x}{5}+\dfrac {sin\;5x}{25}+C\)

B \(\dfrac {x\;cos5x}{5}+C\)

C \(x\, sin\, 5x + x^2 cos\; 5x + C\)

D \(x^3 + x^2 +C\)

- The formula of integration by parts is

\(\int udv=uv-\int vdu\) ... (1)

**Step 1:** First, identify which function is ** u** and which one is **dv**.

**Step 2:** Calculate** du** and** v**.

**Step 3:** Put the values in the formula.

**Step 4:** Evaluate

- To identify
**u**and**dv**, we look at the second part of RHS of equation (1). - We choose
**u**and**dv**in such order so as to make the integral easy to integrated.

For example: 1. \(\int te^t\,dt\)

**Step 1:** Identify** u** and** dv**.

If we choose **e ^{t}** as

Whereas, if we choose** t** as **u** and **e ^{t}dt** as

2. \(\int (\sin x)(\ell n x)\,dx\)

Here, if we choose **sin x** as** u** and **ln x dx** as **dv**, then the second part of RHS of equation (1) will be the integral of three or more functions and difficult to calculate.

Whereas, if we choose** ln x** as **u** and **sinx dx** as **dv**, then the second part of RHS of equation (1) will be \(\int \dfrac{1}{x}\cos x\,dx\) which seems easier than the previous one.

3. \(\int x \sin x\,dx\)

If we choose **sinx** as **u** and** x dx** as** dv**, then the second part of RHS of equation (1) will be \(\displaystyle\int \cos x\left(\dfrac{x^2}{2}\right)\,dx\), which will be complicated.

Whereas, if we choose** x** as** u** and** sinx dx** as **dv**, then the second part of RHS of equation (1) will be \(\int \cos x\,dx\) which is just a trigonometric function that can be calculated easily.

4. \(\int \cos^{-1}x\,dx\)

Here, we do not know the integral of **cos ^{-1}x dx**. So, we will assume the integrand as the product of

If we choose **1** as **u** and **cos ^{-1}x dx** as

So, we will choose **cos ^{-1}x** as

**Conclusion:**

- We should choose
**u**as a function which on differentiation reduces the complexity of \(\int vdu\). - Also, we should choose
**dv**as a function which on integration reduces the complexity of \(\int vdu\). - Let's us solve example (1)

Now, \(I=\int t e^tdt\)

So,** t=u** and **e ^{t}dt=dv**... (2)

**Step 2:** Find **du** and **v**

**Step 3:** Put the values in the formula.

Example: \(\int \ell nx\,dx\)

Generally we know the direct formula for integration of **ln x, **but it can't be integrated directly as a single function.

So, we treat this function as a product of 1 and \(\ell nx\), i.e., \(\int 1\cdot \ell nx \,dx\)

Here, we will derive the formula for integrating the function of **ln x** using integration by parts formula.

**Step 1:** Identify** u** and **dv**.

If we choose **1** as** u** and** ln x dx** as **dv**, then we don't know the integral of **ln x dx**. So, we can't get** v**.

Whereas, if we choose** ln x** as** u** and **1.dx** as **dv**, then the second part of RHS of equation (1) will be \(\displaystyle \int x\left(\dfrac{1}{x}\right)\,dx\) which can be calculated easily.

So here, \(\ell nx=u\) and \(dv=1\cdot dx\)

**Step 2:** Find \(du\) and \(v.\)

**Step 3:** Put the values in the formula,

This is the required integral of\(\int \ell nx\,dx\)

- Let's take another example.

\(\int x^n\ell nx\,dx\)

**Step 1:** Identify \(u\) and \(dv\).

Here, derivative of \(x^n\) is \(nx^{n-1}\)

derivative of \(\ell nx \) is \(\dfrac{1}{x}\)

And anti-derivative of \(x^n\) is \(\dfrac{x^{n+1}}{n+1}\)

anti-derivative of \(\ell nx\) is \(x(\ell nx-1)\)

- If we choose \(x^n\) as \(u\) and \(\ell nx\,dx\) as \(dv\), then the second part of RHS of equation (1) will be \(\int x(\ell nx-1)nx^{n-1}\,dx\).

This is the product of functions which is very complicated to integrate.

Whereas, if we choose \(x^ndx\) as \(dv\) and \(\ell nx\) as \(u,\) then the second part of RHS of equation (1) will be \(\displaystyle \int \dfrac{x^{n+1}}{n+1}\,\dfrac{1}{x}\,dx\) which can be easily calculated.

So, \(u=\ell nx\) and \(x^ndx=dv\)

**Step 2:** Find \(du\) and \(v.\)

**Step 3:** Put the values in the formula-

Trick to choose** u:**

- Many students get confused while choosing \(u\) and \(v.\) So, there is a trick to choose \(u.\)

A \(e^t+sin\;t+C\)

B \(\dfrac {e^{2t}}{4}[2t-1]+C\)

C \(2\;cos^2\;t+sin^2\;t+C\)

D \(\dfrac {e^{2t}}{2}[4t-1]+C\)

Integration by parts is a method to integrate the product of two functions.

The formula for integration by parts is:

\(\int\,u\,dv=uv-\int v\, du\)

The second term on R.H.S. of the formulas tells us that \(u\) is the part that would preferably be differentiated and \(dv\) is the part that would preferably be integrated.

- In some cases, applying integration by parts once will not solve the problem. We might need to apply it more than once.

**Ex. : **To integrate the function of the form:

\(\displaystyle I=\int(\text { Polynomial })\;e^{ax}\;dx\)

where \(a\in R\),

- We know while applying integration by parts method, we choose is
**u**= Polynomial and \(dv=e^{ax}\) . Because differentiating the polynomial gives simpler term in comparison to if we choose \(dv=(\text { Polynomial })\;dx\), then its integration will lead to higher power term, where as integration of \(e^{ax}\) remains the same.

**Ex.: **If \(u=x^2\) and if \(dv=x^2\;dx\)

\(\dfrac {du}{dx}=2x\) \(v=\dfrac {x^3}{3}\)

- Thus, \(u=\) Polynomial and \(dv=e^{x}\;dx\) is a good choice when we are integrating product of a polynomial and an exponential function.
- On differentiating a polynomial function of degree \(n\) one time, its power reduces by 1 and we get a polynomial function of degree \(n-1\),

**Ex: **Let \(u=x^3\), if we differentiate it once, we get \(du=3x^2\;dx\)

We observe that here power \(3\) reduces to \(2\) on differentiating it once.

So, we can conclude that to get the answer, integration by parts method is performed as many times as the degree of polynomial.

**Ex. \(\int\;x^2\,e^x\;dx\)**

**\(u=x^2,\;\;\;\;\;\;\;dv=e^x\;dx\)**

\(du=2x\,dx,\;\;\;\;v=e^x\)

On applying integration by parts formula once,

Again we get the product of two functions. So, once again, we apply integration by parts formula.

Let \(I_1=\int x\,e^x \;dx\)

\(u=x\) \(dv=e^x\;dx\)

\(du=dx\) \(v=e^x\)

So,

\(I_1=x\,e^x-\int e^x \;dx\)

\(I_1=x\,e^x-e^x\) .... (ii)

Now, put (ii) in (i)

So,

- In the above example, we saw that the degree of polynomial was 2 (highest power of \(x\)), and we also used the integration by parts, method two times.
- When we have product of two functions will cyclic derivative, then also integration by parts is used more than once.

**Ex.: \(\int \,e^x\,cos\,x\;dx\) **or \(\int \,e^x\,sin\,x\;dx\)

- If we have product of powers of polynomial and logarithmic functions, then also integration by parts is used more than once.

**Note: **For higher powers of \(x\) of the form,

\(\int \,x^x\,e^x\,\;dx,\;\int \,x^x\,sin\,x\;dx,\;\int \,x^x\,cos\,x\;dx\)

repeatedly use of integration by parts can evaluate the integrals. Each time, the method reduces the power of \(x\) by one.

A \(e^{2x}[x^2+3x+4]+C\)

B \(\dfrac {e^{2x}}{4}[2x^2-2x+7]+C\)

C \(sin\, x + log \,(tan\, x) + C\)

D \(\dfrac {e^{2x}}{4}[x^2-4x+7]+C\)

- In some indefinite integrals, first we make an appropriate substitution and then the new integral formed, can be integrated by method of integration by parts.

For e.g.,

Consider

\(I=\int e^{\sqrt x}\;dx\)

If we put \(x = t^2\), then \(dx = 2t\; dt\)

\(I=\int e^t×2t\;dt\)

\(I=2\int t\,e^t\;dt\) (This will use integration by parts, so first we use substitution and then integration by parts)

A \(e^{sin\,x}×\sqrt x+C\)

B \(-2\sqrt {x}\;\;sin\sqrt {x}\;+2cos\;\sqrt {x}\;+C\)

C \(e^{cos\,x}×\sqrt {x}\;+C\)

D \(-2\sqrt {x}\;\;cos\sqrt {x}\;+2\;sin\;\sqrt {x}\;+C\)

A \(\tan x+\ell n(1+x^2)+C\)

B \(\tan ^{-1}x+\dfrac{1}{2}\ell n(x^2)+C\)

C \(\tan^{-1}x-\ell n(1+x^2)\)

D \(x \tan^{-1}x-\dfrac{1}{2}\ell n(1+x^2)+C\)

- Some integrals don't give an answer on applying integration by parts method once.
- In those cases, we apply the method more than once.
- There are some cases, where the original integral on the right hand side of integration by parts formula reappears.
- This type of problem occurs when the product of two functions involve cyclic derivative (since cosine etc.)
- We know that on differentiating sine and cosine functions two times, we again get the same functions, we and cosine respectively.

For** Ex.:** \(u=sin \,x\)

(1) \(\dfrac {du}{dx}=cos \,x\)

(2) \(\dfrac {d^2u}{dx^2}=-sin \,x\)

- Let us understand by taking another example:

\(I=\int\, e^{ax}\,cos\,bx\;dx\)

- Since the above given integral is a product of two functions, so we use integration by parts method.

**Conclusion:**

- We should choose \(u\) as a function which on differentiation reduce the complexity of \(\int v\;du\)
- Also, we should choose \(dv\) as a function which on integration reduces the complexity of \(\int v\;du\).

So,

\(u=cos\,bx\) and \(dv=e^{ax}\;dx\)

\({u'=-b\,sin\,bx}\) \(v=\dfrac {e^{ax}}{a}\)

Now, applying integration by parts formula.

\(\int u\,dv=u\,v- \int v\,du\)

\(I=\int e^{ax}\,cos\,bx\,dx=cos\,bx×\dfrac {e^{ax}}{a}-\displaystyle\int\dfrac {e^{ax}}{a}×-b\,sin\,bx\;dx\)

Again we get the product of two functions, so we again, we integration by parts

\(I_1=\int e^{ax}\,sin\,bx\;dx\)

Again, let \(u=sin\,bx\) , \(dv=e^{ax}\;dx\)

\(u'=b\,cos\;bx\) , \(v=\dfrac {e^{ax}}{a}\)

\(I_1=\dfrac {e^{ax}}{a}\,sin\,bx- \displaystyle\int \dfrac {e^{ax}}{a}× b\,cos\,bx\;dx\)

\(I_1=\dfrac {e^{ax}}{a}\,sin\,bx-\dfrac {b}{a} \displaystyle\int e^{ax}×\,cos\,bx\;dx\)

Here, we can use that \(\displaystyle\int e^{ax} \,cos\,bx\;dx=I\) appears again,

So,

\(I_1=\dfrac {e^{ax}}{a}\,sin\,bx-\dfrac {b\,I}{a}\) ... (ii)

Put (ii) in (i),

\(I_1=\dfrac {e^{ax}}{a}\,cos\,bx+\dfrac {b}{a} \left[ \dfrac {e^{ax}}{a} sin\,bx-\dfrac {b}{a} I\right]\)

\(I_1=\dfrac {e^{ax}}{a}\,cos\,bx+\dfrac {b}{a} e^{ax} sin\,bx-\dfrac {b^2}{a^2} I\)

\(I+\dfrac {b^2}{a^2} I= \dfrac{ a\,e^{ax}\,cos\,bx+b\,e^{ax}\,sin\,bx}{a^2}\)

\(\dfrac {a^2I+b^2I}{a^2}= \dfrac{ a\,e^{ax}\,cos\,bx+b\,e^{ax}\,sin\,bx}{a^2}\)

\(I(a^2+b^2)= e^{ax}[a\,cos\,bx+b\,sin\,bx]\)

\(I=\dfrac { e^{ax}[a\,cos\,bx+b\,sin\,bx] } {(a^2+b^2)}\)

\(\Rightarrow \displaystyle\int e^{ax}\,cos\,bx\,dx=\dfrac {e^{ax}}{(a^2+b^2)}[a\, cos\, bx + b\, sin\, bx]\)

Similarly,

\(\Rightarrow \displaystyle\int e^{ax}\,sin\,bx\,dx= \dfrac {e^{ax}}{(a^2+b^2)}[a\, sin\, bx - b\, cos\, bx]\)

A \(\dfrac {e^{2x}}{5}(2\,cos\,x-sin\,x)+C\)

B \(e^{x}\,sin\,x+\ell nx+C\)

C \(\dfrac {e^{2x}}{5}(2\,sin\,x-cos\,x)+C\)

D \(\ell n\,(sin\,x)+e^{x}+C\)

- If we combine the formula for Integration by Parts with Fundamental Theorem of Calculus II, we can find the definite integral by parts.

\(\int f(x)\,g'(x)\;dx=f(x)\;g(x)-\int g(x)\,f'(x)\;dx\)

\(\int\limits_a^b\;f(x)\;g'(x)\;dx=f(x)\,g(x)\Big]_a^b-\int\limits_a^bg(x)\;f'(x)\;dx\)

where both the terms of R.H.S should be finite.

A \(\dfrac {1}{\ell n\,2}\)

B \(\dfrac {2}{3}\,e\)

C \(\dfrac {1}{2}\,{\ell n} \left (\dfrac {2}{e} \right)\)

D \(\dfrac {1}{2}\,{\ell n} \left (\dfrac {e}{2} \right)\)

- Integration by parts method is used to integrate the product of two functions.
- To evaluate the integral involving product of the power of polynomial and logarithmic function, we first need to identify
**u**and**dv**. - Let us understand by taking an example:

\(I=\int x^m (log\,x)^n\;dx\)

**Selection of u and dv:**

We know that integration by parts formula is:

\(\int u\,dv=uv-\int v\,du\)

Now if we take **u=x**** ^{m }**and

Whereas, if we take, **u=(log x) ^{n}** and

\(\dfrac {d}{dx}\,log\,x=\dfrac {1}{x}\).

So, \(u=(log\,x)^n\) and \(dv=x^m\;dx\) is a good choice.

But to obtain the answer, we have to differentiate \((log\,x)^n\) again and again till we get 1 out of it. We know that each time on differentiation the power of \((log\,x)^n\) reduces by one.

\(\dfrac {d}{dx}(log\,x)^n=n (log\,x)^{n-1}×\dfrac {1}{x}\)

So, to evaluate this integral, we have to apply integration by parts as many times as the number **n**.

**Ex.:**

**\(I=\int x^4 (\ell n x)^3\;dx\)**

So,

**\(u=(\ell n\,x)^3\) **and \(dv=x^4\;dx\)

\(du=\dfrac {3}{x}(\ell n\,x)^2\;dx\) \(v=\dfrac {x^5}{5}\)

By integration by parts formula, we have

Again we obtain product of two functions, so by integration by parts formula, we have,

\(I_1=\int x^4 (\ell n x)^2\;dx\)

Let

Now, again on applying integration by parts formula, we have,

\(I_2=\int x^4\,\ell n\,x\;dx\)

Let

Putting (iii) in (ii)

\(I_1=\dfrac {x^5}{5} (\ell nx)^2-\dfrac {2}{5} \left [ \dfrac {x^5}{5}\;\ell n\,x\;-\dfrac {x^5}{25} \right]\)... (iv)

Putting (iv) on (i)

A \(\dfrac {x^4}{4}(\ell n\,x)^2-\dfrac {x^4}{8}\ell n\,x+\dfrac {x^4}{32}+C\)

B \(\dfrac {x^4}{8}(\ell n\,x)^2-\dfrac {x^4}{4}\ell n\,x+x^4+C\)

C \(x^3\,e^x+C\)

D \(x\,\ell n\,x+C\)