Standard Substitutions in Indefinite Integral - Solved Examples

The indefinite integral is a primary topic in integral calculus. It is the foundation for definite integrals. This article gives a clear idea about standard substitutions in indefinite integral.

The use of integration can be seen in the following kinds of problems:

A) The first kind of problem deals with those where the derivative of the function or slope is known and the function is unknown. Hence, integration (ani-differentiation) has to be used in this case.

B) The second kind of problem is definite integrals. To find the area, volume, moment of inertia and work done by a force, definite integrals are used.

An integral can be called an indefinite integral if the value is not determined fully till the endpoints are defined. This vagueness is dealt with by the addition of the constant of integration C at the end. The constant C is added because there are infinite solutions to the integral.

Suppose a function f is there, and we need to find a function F, such that the derivative of F is equal to f, then F is the indefinite integral of f. It is written as

\(\begin{array}{l}F(x)=\int f(x)dx\end{array} \)

Where f(x) is the integrand, x is the variable of integration, and C is the constant of integration.

The 3 methods of solving indefinite integrals are integration by parts, integration by substitution, and integration by partial fractions.

This article discusses integration by standard substitution of indefinite integrals.

The substitution method comprises two parts, namely direct and indirect substitution. However, integrals can be solved by standard substitution also. The following are some of the standard substitution formulae:

If terms are of the type x² + a² or √[x² + a²], set x = a tan θ or a cot θ
If terms are of the type x² – a² or √[x² – a²], set x = a sec θ or a cosec θ
If terms are of the type a² – x² or √[x² + a²], set x = a sin θ or a cos θ
If √a + x, √a – x, are there, then set x = a cos θ.
For the kind √(x – a) (b – x), set x = a cos2θ + b sin2θ
For the kind (√x² + a² ± x)ⁿ or (x ± √x² – a²)ⁿ, set the term in the bracket = t.
For 1 /(x + a)ⁿ¹ (x + b)ⁿ², n1, n2 ∈ N (and > 1), again set (x + a) = p(x + b)

Also Read

Definite and Indefinite Integration

Important Integration Formulas

Standard Substitutions in Indefinite Integral Examples

Example 1: Find ∫x dx / [1 − x cotx].

Solution:

∫x dx / [1 − x cotx] = ∫ x dx / (1−x) * [cosx / sinx]

= ∫(x sinx / [sinx − x cosx]) dx

= ∫dt / t = logt = log (sinx − x cosx) + c

{By sinx − x cosx = t [cosx − ( −x sinx + cosx)] dx = dt⇒ x sinx dx = dt}

Example 2: ∫[x³ / √x²+ 2] dx

Solution:

∫[x³ / √x²+ 2] dx =∫[x²* x / √x²+ 2] dx

Put x²+ 2 = t²⇒x dx = t dt and x²= t²− 2, then it reduces to

∫[(t²− 2) / t ] t dt =∫(t²− 2)dt

= [t³/ 3] − 2t + c = [(x²+ 2 )^3/2/ 3] −2 (x²+ 2)^½+ c

Example 3: ∫[e^x(x + 1)] / cos²(xe^x) dx

Solution:

∫[e^x(x + 1)] / cos²(xe^x) =∫e^x(x + 1) sec²(xe^x) dx

Putting xe^x= t ⇒ (x + 1) e^xdx = dt, we get

∫sec²t dt = tant + c = tan (xe^x) + c

Example 4: ∫(√tanx/ [sinx cosx]) dx

Solution:

∫(√tanx/ [sinx cosx]) dx =∫[tanx] / [√(tanx) * sinx* cosx] dx

=∫[sinx * secx] / [√(tanx) * sinx* cosx] dx

=∫[sec²x / √(tanx)] dx

Put t = tanx ⇒ dt = sec²x dx, then it reduces to

∫[1 / √t] dt = 2t^1/2+ c = 2√tanx + c

Example 5: ∫e^−x/ [1 + e^x] dx

Solution:

∫e^−x/ [1 + e^x] dx = ∫([e^−x* e^−x ] / [e^−x + 1]) dx

Put e^−x+ 1 = t ⇒ −e^−xdx = dt, then it reduces to

−∫[(t−1) / t ] dt =∫{([1 / t] −1)} dt

= log t − t + c

= log(e^−x+ 1) − (e^−x+ 1) + c

= log (e^x+ 1) − x − e^−x− 1 + c

= log (e^x+ 1) − x − e^−x− 1 + c, (∵1 = constant)

Example 6: ∫(cosx √[4−sin²x]) dx

Solution:

∫(cosx √[4−sin²x]) dx

Putting sinx = t ⇒ cosx dx = dt, we get

∫(cosx √[4−sin²x]) dx =∫√[4 − t²] dt =∫[√[(2)²− t²] dt

= (t / 2) √[4 − t²] + (4 / 2) sin⁻¹ (t / 2) + c

= (1 / 2) sinx √[4 − sin²x] + 2sin⁻¹[(1 / 2) sinx)] + c

Example 7:

\(\begin{array}{l}\int_{{}}^{{}}{\frac{\sec x\ dx}{\sqrt{\cos 2x}}}=\end{array} \)

Solution:

\(\begin{array}{l}\int_{{}}^{{}}{\frac{\sec x\,dx}{\sqrt{\cos 2x}}}=\int_{{}}^{{}}{\frac{\sec x}{\sqrt{{{\cos }^{2}}x-{{\sin }^{2}}x}}}\,dx \\ =\int_{{}}^{{}}{\frac{{{\sec }^{2}}x\,dx}{\sqrt{1-{{\tan }^{2}}x}}}\end{array} \)

[Multiplying numerator and denominator by secx]

Now, putting tanx = t ⇒ sec²x dx = dt, we get the integral

\(\begin{array}{l}={{\sin }^{-1}}t={{\sin }^{-1}}(\tan x).\end{array} \)

Trick : Since

\(\begin{array}{l}\frac{d}{dx}\{{{\sin }^{-1}}(\tan x)\}=\frac{{{\sec }^{2}}x}{\sqrt{1-{{\tan }^{2}}x}} =\frac{{{\sec }^{2}}x.\cos x}{\sqrt{{{\cos }^{2}}x-{{\sin }^{2}}x}}=\frac{\sec x}{\sqrt{\cos 2x}}.\end{array} \)

Example 8:

\(\begin{array}{l}\int_{{}}^{{}}{\frac{\sin 2x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx=}\end{array} \)

Solution:

\(\begin{array}{l}\int_{{}}^{{}}{\frac{\sin 2x}{{{\sin }^{4}}x+{{\cos }^{4}}x}\,dx} \\ =\int_{{}}^{{}}{\frac{2\sin x\cos x}{{{\sin }^{4}}x+{{\cos }^{4}}x}\,dx=\int_{{}}^{{}}{\frac{2\tan x{{\sec }^{2}}x}{1+{{\tan }^{4}}x}\,dx}}\end{array} \)

Put tan²x = t ⇒2 tanx sec²x dx = dt, we get;

\(\begin{array}{l}\Rightarrow f(x)=\frac{{{x}^{3}}}{3}+5x+c\end{array} \)

Trick:

\(\begin{array}{l}\frac{d}{dx}\left\{ {{\cot }^{-1}}({{\tan }^{2}}x) \right\}=-\frac{1(2\tan x\,.\,{{\sec }^{2}}x)}{1+{{\tan }^{4}}x}=-\frac{\sin 2x}{{{\cos }^{4}}x+{{\sin }^{4}}x} \\ \Rightarrow \frac{d}{dx}\left\{ {{\tan }^{-1}}({{\tan }^{2}}x) \right\}=\frac{\sin 2x}{{{\sin }^{4}}x+{{\cos }^{4}}x}\end{array} \)

Example 9:

\(\begin{array}{l}\int_{{}}^{{}}{\frac{x+1}{\sqrt{1+{{x}^{2}}}}dx}=\end{array} \)

Solution:

\(\begin{array}{l}\int_{{}}^{{}}{\frac{x+1}{\sqrt{{{x}^{2}}+1}}\,dx=\int_{{}}^{{}}{\frac{x}{\sqrt{{{x}^{2}}+1}}\,dx+\int_{{}}^{{}}{\frac{1}{\sqrt{{{x}^{2}}+1}}\,dx}}}\end{array} \)

Put x² + 1 = t ⇒2x dx = dt, then it is reduced to

\(\begin{array}{l}\frac{1}{2}\int_{{}}^{{}}{\frac{dt}{{{t}^{1/2}}}+\int_{{}}^{{}}{\frac{1}{\sqrt{{{x}^{2}}+1}}}}\,dx=\frac{1}{2}.2.{{t}^{1/2}}+\log (x+\sqrt{{{x}^{2}}+1})+c \\={{({{x}^{2}}+1)}^{1/2}}+\log (x+\sqrt{{{x}^{2}}+1})+c\end{array} \)

Example 10:

\(\begin{array}{l}\int_{{}}^{{}}{\frac{1}{\sqrt{1-{{e}^{2x}}}}\ dx=}\end{array} \)

Solution:

\(\begin{array}{l}\int_{{}}^{{}}{\frac{1}{\sqrt{1-{{e}^{2x}}}}\,dx=\int_{{}}^{{}}{\frac{{{e}^{-x}}}{\sqrt{{{e}^{-2x}}-1}}\,dx}}\end{array} \)

Put e^-x = t ⇒ -e^-x dx = dt, then it is reduced to

\(\begin{array}{l}-\int_{{}}^{{}}{\frac{1}{\sqrt{{{t}^{2}}-1}}\,dt=-\log \left[ t+\sqrt{{{t}^{2}}-1} \right]+c} \\ =-\log \left[ {{e}^{-x}}+\sqrt{{{e}^{-2x}}-1} \right]=-\log \left[ \frac{1}{{{e}^{x}}}+\frac{\sqrt{1-{{e}^{2x}}}}{{{e}^{x}}} \right] \\ =-\log \left[ 1+\sqrt{1-{{e}^{2x}}} \right]+\log {{e}^{x}}+c \\ =x-\log \left[ 1+\sqrt{1-{{e}^{2x}}} \right]+c.\end{array} \)

Example 11:

\(\begin{array}{l}\int_{{}}^{{}}{\frac{3{{x}^{2}}}{\sqrt{9-16{{x}^{6}}}}}\ dx=\end{array} \)

Solution:

\(\begin{array}{l}\int_{{}}^{{}}{\frac{3{{x}^{2}}}{\sqrt{9-16{{x}^{6}}}}dx=\int_{{}}^{{}}{\frac{3{{x}^{2}}}{\sqrt{{{(3)}^{2}}-{{(4{{x}^{3}})}^{2}}}}\,dx}}\end{array} \)

Put 4x³ = t ⇒ 12x² dx = dt, then it is reduced to

\(\begin{array}{l}\frac{1}{4}\int_{{}}^{{}}{\frac{dt}{\sqrt{{{(3)}^{2}}-{{t}^{2}}}} \\ =\frac{1}{4}.\frac{1}{1}{{\sin }^{-1}}\left( \frac{t}{3} \right)+c \\ =\frac{1}{4}{{\sin }^{-1}}\left( \frac{4{{x}^{3}}}{3} \right)}+c\end{array} \)

Indefinite Integration – Video Lesson

Indefinite Integration - Video Lesson

Indefinite Integration Problems

Indefinite-Integration Problems

Frequently Asked Questions

Q1

What is the use of the substitution method in integration?

When we do a substitution for a function whose derivative also exists in the integrand, the substitution method is very useful. This makes the function simplified, and we can use the basic formulas of integration to integrate the function.

Q2

Give the substitution rule for integration.

∫ f(g(x)) g’(x) dx = ∫f(u) du, where u = g(x), du = g’(x).

Q3

Find ∫[(log x)/x] dx.

We have to find ∫[(log x)/x ]dx. Put u = log x, then du = (1/x) dx.
So ∫(log x)/x dx = ∫u du = (u²/2) + c
= [(log x)²/2] + c

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