Standard Substitutions in Indefinite Integral

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 problems dealing with those where the derivative of 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 problems are about the definite integrals. To find the area, volume, the moment of inertia, 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,

F(x) = F(x)=f(x)dxF(x)=\int f(x)dx

where f(x) is 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, 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: ∫[ex (x + 1)] / cos2 (xex) dx

Solution:

∫[ex (x + 1)] / cos2 (xex) =∫ex (x + 1) sec2 (xex) dx

Putting xex = t ⇒ (x + 1) ex dx = dt, we get

∫sec2 t dt = tant + c = tan (xex) + c

Example 4: ∫(√tanx/ [sinx cosx]) dx

Solution:

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

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

=∫[sec2 x / √(tanx)] dx

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

∫[1 / √t] dt = 2t1/2 + c = 2√tanx + c

Example 5: ∫e−x / [1 + ex] dx

Solution:

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

Put e−x + 1 = t ⇒ −e−x dx = 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 (ex + 1) − x − e−x − 1 + c

= log (ex + 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: secx dxcos2x=\int_{{}}^{{}}{\frac{\sec x\ dx}{\sqrt{\cos 2x}}}=

Solution:

secxdxcos2x=secxcos2xsin2xdx=sec2xdx1tan2x\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}}} [Multiplying numerator and denominator by secx]

Now putting tanx=tsec2xdx=dt,\tan x=t\Rightarrow {{\sec }^{2}}x\,dx=dt, we get the integral  

=sin1t=sin1(tanx).={{\sin }^{-1}}t={{\sin }^{-1}}(\tan x).

Trick : Since ddx{sin1(tanx)}=sec2x1tan2x=sec2x.cosxcos2xsin2x=secxcos2x.\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}}.

Example 8: sin2xsin4x+cos4xdx=\int_{{}}^{{}}{\frac{\sin 2x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx=}

Solution: 

sin2xsin4x+cos4xdx=2sinxcosxsin4x+cos4xdx=2tanxsec2x1+tan4xdx\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}}

Put tan2x=t2tanx sec2xdx=dt,{{\tan }^{2}}x=t\Rightarrow 2\tan x\ {{\sec }^{2}}x\,dx=dt,

then it is reduced to f(x)=x33+5x+c\Rightarrow f(x)=\frac{{{x}^{3}}}{3}+5x+c

Trick: ddx{cot1(tan2x)}=1(2tanx.sec2x)1+tan4x=sin2xcos4x+sin4xddx{tan1(tan2x)}=sin2xsin4x+cos4x\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}

Example 9: x+11+x2dx=\int_{{}}^{{}}{\frac{x+1}{\sqrt{1+{{x}^{2}}}}dx}=

Solution: 

x+1x2+1dx=xx2+1dx+1x2+1dx\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}}}

Put x2+1=t2xdx=dt{{x}^{2}}+1=t\Rightarrow 2x\,dx=dt, then it is reduced to 

12dtt1/2+1x2+1dx=12.2.t1/2+log(x+x2+1)+c=(x2+1)1/2+log(x+x2+1)+c\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

Example 10: 11e2x dx=\int_{{}}^{{}}{\frac{1}{\sqrt{1-{{e}^{2x}}}}\ dx=}

Solution: 

11e2xdx=exe2x1dx\int_{{}}^{{}}{\frac{1}{\sqrt{1-{{e}^{2x}}}}\,dx=\int_{{}}^{{}}{\frac{{{e}^{-x}}}{\sqrt{{{e}^{-2x}}-1}}\,dx}}

Put ex=texdx=dt,{{e}^{-x}}=t\Rightarrow -{{e}^{-x}}dx=dt, then it is reduced to 

1t21dt=log[t+t21]+c=log[ex+e2x1]=log[1ex+1e2xex]=log[1+1e2x]+logex+c=xlog[1+1e2x]+c.-\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.

Example 11: 3x2916x6 dx=\int_{{}}^{{}}{\frac{3{{x}^{2}}}{\sqrt{9-16{{x}^{6}}}}}\ dx=

Solution: 

3x2916x6dx=3x2(3)2(4x3)2dx\int_{{}}^{{}}{\frac{3{{x}^{2}}}{\sqrt{9-16{{x}^{6}}}}dx=\int_{{}}^{{}}{\frac{3{{x}^{2}}}{\sqrt{{{(3)}^{2}}-{{(4{{x}^{3}})}^{2}}}}\,dx}}

Put 4x3=t12x2dx=dt,4{{x}^{3}}=t\Rightarrow 12{{x}^{2}}dx=dt, then it is reduced to 

14dt(3)2t2=14.11sin1(t3)+c=14sin1(4x33)+c\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

BOOK

Free Class