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Integration By Substitution Of Trigonometric Functions

Learn with the example of using trig substitution to evaluate an indefinite integration. Practice integration by completing the squares and calculus trigonometric substitution.

Inverse Substitutions

  • Sometimes, we come across irrational terms like \(\sqrt{a^2-x^2},\,\sqrt{a^2+x^2}\) in the integrand function to remove the radical sign we make some trigonometric substitution.
  • We make substitutions of the form  \(x=g(t)\) or \(g(\theta)\,\to\) this reverse substitution is called inverse substitution.

 Some standard substitutions are

  1. \(\sqrt{a^2-x^2}\,\to\) put  \(x=a\,sin\,\theta\)
  2. \(\sqrt{x^2+a^2}\,\to\) put \(x=a\,tan\,\theta\)
  3. \(\sqrt{x^2-a^2}\,\to\) put \(x=a\,sec\,\theta\)
  4. \(\sqrt{\dfrac{a-x}{a+x}}\,\to\) put \(x=a\,cos\,2\theta\)

When we make these substitution, we get a perfect square under the radical signs. This simplifies the integrand function.

If the integral is definite then, when we make the substitution we change the limit according to new variables so that we do not have to go back to original variable.

Illustration Questions

Identify the appropriate substitution for the given integral \(\displaystyle I=\int\dfrac{\sqrt {9-x^2}}{x}dx\)

A \(x=3sec\,\theta\)

B \(x=tan\theta\)

C \(x=3\,sin\theta\)

D \(x=cot\theta\)

×

The integrand contains the expression \(\sqrt{9-x^2}\) which is of the form \(\sqrt{a^2-x^2}\)  

when  \(a=3\)

Make the substitution 

\(x=3sin\theta\)

\(\therefore\, \) Option 'C' is correct.

Identify the appropriate substitution for the given integral \(\displaystyle I=\int\dfrac{\sqrt {9-x^2}}{x}dx\)

A

\(x=3sec\,\theta\)

.

B

\(x=tan\theta\)

C

\(x=3\,sin\theta\)

D

\(x=cot\theta\)

Option C is Correct

Trigonometric Substitution by Tangent Function

  • Sometimes we come across irrational terms like \(\sqrt{a^2-x^2},\,\sqrt{a^2+x^2}\) in the integrand function to remove the radical sign we make some trigonometric substitution.
  • We make substitution of the form  \(x=g(t)\) or \(g(\theta)\,\to\) This reverse substitution is called inverse substitution.

If the integrand contains the expression \(\sqrt{a^2+x^2}\) where 'a' is a constant, we make the substitution \(x=a\,tan\,\theta\)\(\left(-\dfrac{\pi}{2}<\theta<\dfrac{\pi}{2}\right)\) convert the integrand in terms of  \(\theta\), then evaluate.

Illustration Questions

Evaluate \(\displaystyle I=\int\dfrac{\sqrt {1+x^2}}{x^4}dx\)

A \(-\dfrac{1}{3}\,\dfrac{(1+x^2)^{3/2}}{x^3}+C\)

B \(-\dfrac{1}{3}\,\dfrac{(1+x^2)^{3/2}}{x^4}+C\)

C \(-\sqrt{1+x^2}+x+C\)

D \(x^2ln\,x+sin\,x+C\)

×

\(\displaystyle I=\int\dfrac{\sqrt {1+x^2}}{x^4}dx\)

Since \(\sqrt{1+x^2}\) is present in the integrand

put, \(x=tan\,\theta\)

\(\Rightarrow \,dx=sec^2\theta \,d\theta\)

\(\therefore\, \) \(\displaystyle I=\int\dfrac{\sqrt {1+tan^2\theta}×sec^2\theta\, d\theta}{tan^4\theta}\)

\(=\displaystyle I=\int\dfrac{|sec\,\theta|\,sec^2\theta\, d\theta}{tan^4\theta}\)

\(|sec\,\theta|=sec\,\theta\) as \(-\dfrac{\pi}{2}\leq\theta\leq\dfrac{\pi}{2}\)

\(=\displaystyle \int\dfrac{sec^3\theta}{tan^4\theta}\)

\(=\displaystyle \int\dfrac{cos\,\theta}{sin^4\theta}d\theta\)

Put \(sin\,\theta=t\)

\(\Rightarrow\,cos\,\theta\,d\theta=dt\)

\(\Rightarrow\,\displaystyle I=\int\dfrac{dt}{t^4}\\=\dfrac{t^{-3}}{-3}+C\)

\(\Rightarrow\,\dfrac{-1}{3sin^3\theta}+C\)

Now, we go back to original variable  \(x\)

\(x=tan\,\theta\)

\(sin\theta=\dfrac{x}{\sqrt{1+x^2}}\)

\(\therefore\,I=-\dfrac{1(1+x^2)^{3/2}}{3x^3}\)

\(=-\dfrac{1}{3}\dfrac{(1+x^2)^{3/2}}{x^3}+C\)

image

Evaluate \(\displaystyle I=\int\dfrac{\sqrt {1+x^2}}{x^4}dx\)

A

\(-\dfrac{1}{3}\,\dfrac{(1+x^2)^{3/2}}{x^3}+C\)

.

B

\(-\dfrac{1}{3}\,\dfrac{(1+x^2)^{3/2}}{x^4}+C\)

C

\(-\sqrt{1+x^2}+x+C\)

D

\(x^2ln\,x+sin\,x+C\)

Option A is Correct

Trigonometric Substitution by Secant Function

  • Sometimes we come across irrational terms like \(\sqrt{a^2-x^2},\,\sqrt{a^2+x^2}\) in the integrand function to remove the radical sign we make some trigonometric substitution.
  • We make substitution of the form  \(x=g(t)\) or \(g(\theta)\,\to\) this reverse substitution is called inverse substitution.

If the integrand contain the expression \(\sqrt{x^2-a^2}\) where, 'a' is a constant, we make the substitution \(x=a\,sec\,\theta\)  \(\left(0\leq\theta\leq\dfrac{\pi}{2} \,or\,\pi\le\theta\leq\dfrac{3\pi}{2}\right)\) convert the integrand to \(\theta\) and then evaluate.

Illustration Questions

Evaluate \(\displaystyle I=\int\dfrac{dx}{x^2\sqrt {x^2-9}}\)

A \(\dfrac{\sqrt{x^2-9}}{9x}+C\)

B \(x^2sinx+lnx+C\)

C \(\dfrac{\sqrt {x^2-9}}{x^2}+C\)

D \(xe^x-\dfrac{2}{x}+C\)

×

\(\displaystyle I=\int\dfrac{1}{x^2\sqrt {x^2-9}}dx\)

Since, \(\sqrt{x^2-9}\) is present we make the substitution

\(x=3\, sec\,\theta\)

\(\Rightarrow \,dx=3\,sec\,\theta \,tan\,\theta\)

\(\therefore\, \) \(\displaystyle I=\int\dfrac{3sec\,\theta \,tan\,\theta\,d\theta}{9sec^2\theta×\sqrt{9sec^2\theta-9}}\)

\(=\displaystyle \dfrac{1}{9}\int\dfrac{sec\,\theta\,tan\,\theta \,d\theta}{sec^2\theta\, tan\,\theta}\)

\(|tan\,\theta|=tan\,\theta\) as \(0\leq\theta\leq\dfrac{\pi}{2}\,or\,\pi<\theta<\dfrac{3\pi}{2}\)

\(=\displaystyle \dfrac{1}{9}\int\dfrac{sec\,\theta\,tan\,\theta\,d\theta}{sec^2\theta \,tan\,\theta}\)

\(=\displaystyle \dfrac{1}{9}\int cos\,\theta\,d\theta\)

\(I=\displaystyle \dfrac{1}{9}\int cos\,\theta\,d\theta\)

\(=\displaystyle \dfrac{1}{9} sin\,\theta+C\)

Now, we go back to the original variable \(x\).

\(x=3\,sec\,\theta\)

\(sin\,\theta=\dfrac{\sqrt{x^2-9}}{x}\)

\(\therefore \,I=\dfrac{\sqrt{x^2-9}}{9x}+C\)

image

Evaluate \(\displaystyle I=\int\dfrac{dx}{x^2\sqrt {x^2-9}}\)

A

\(\dfrac{\sqrt{x^2-9}}{9x}+C\)

.

B

\(x^2sinx+lnx+C\)

C

\(\dfrac{\sqrt {x^2-9}}{x^2}+C\)

D

\(xe^x-\dfrac{2}{x}+C\)

Option A is Correct

Illustration Questions

Find the value of \(\displaystyle I=\int\limits^1_0\ x^2 \sqrt {1-x^2}dx\)

A \(\dfrac{1}{2}\)

B \(\dfrac{\pi}{16}\)

C \(\dfrac{\pi}{8}\)

D \(\dfrac{3\pi}{4}\)

×

\(\displaystyle I=\int\limits^1_0\ x^2 \sqrt {1-x^2}\,dx\)

Since, \(\sqrt{1-x^2}\) is present, we put 

\(x=sin\,\theta\)

\(\Rightarrow\,dx=cos\,\theta\,d\theta\)

Now, when  \(x=1\)

\(\to\theta=\dfrac{\pi}{2}\)

when \(x=0\)

\(\to \theta=0\)

\(\displaystyle\therefore\,I=\int\limits^{{\pi}/{2}}_0sin^2\theta×\sqrt{1-sin^2\theta}×cos\,\theta\,d\theta\)

\(\displaystyle I=\int\limits^{{\pi}/{2}}_0sin^2\theta×|cos\,\theta|cos\,\theta\,d\theta\)

\(|cos\,\theta|=cos\,\theta\) as \(0\leq\theta\leq\dfrac{\pi}{2}\)

\(\displaystyle \Rightarrow\,I=\int\limits^{{\pi}/{2}}_0sin^2\theta×cos^2\theta\,d\theta\)

\(\displaystyle \Rightarrow\,I=\dfrac{1}{4}\int\limits^{{\pi}/{2}}_04×sin^2\theta×cos^2\theta\,d\theta\)

\(\displaystyle \Rightarrow\,I=\dfrac{1}{4}\int\limits^{{\pi}/{2}}_0(2×sin\theta×cos\theta\,)^2d\theta\)

\(\Rightarrow\,\displaystyle \dfrac{1}{4}\int\limits^{{\pi}/{2}}_0sin^2(2\theta)d\theta\)

\(\therefore\,\displaystyle I=\dfrac{1}{4}\,\int\limits^{{\pi}/{2}}_0\dfrac{1-cos\,4\theta}{2}\,d\theta\)

\(=\dfrac{1}{8}\Big[\theta-\dfrac{sin\;4\theta}{4}\Big]^{\dfrac{\pi}{2}}_0\)

\(=\dfrac{1}{8}\left [ \left ( \dfrac {\pi}{2}-0 \right)- \left ( \dfrac {sin 4\; (\pi/2)}{4}-\dfrac {sin0}{4} \right) \right ]\)

\(=\dfrac{1}{8}\left [ \dfrac {\pi}{2}- \left \{ \dfrac {sin \; (2\pi)}{4}-0 \right\} \right ]\)

\(=\dfrac{1}{8}\left [ \dfrac {\pi}{2}- \left \{ 0-0 \right\} \right ]\)          \(\because \;\;sin(2\pi)=0\)

\(=\dfrac{1}{8}\left [ \dfrac {\pi}{2} \right ]\)

\(=\dfrac {\pi}{16} \)

 

Find the value of \(\displaystyle I=\int\limits^1_0\ x^2 \sqrt {1-x^2}dx\)

A

\(\dfrac{1}{2}\)

.

B

\(\dfrac{\pi}{16}\)

C

\(\dfrac{\pi}{8}\)

D

\(\dfrac{3\pi}{4}\)

Option B is Correct

Finding Integrals by Completing the Squares and then using Trigonometric Substitution

  • Consider as integral 

\(\displaystyle I=\int\dfrac{x}{\sqrt{x^2+3x+2}}dx\)

Before we make trigonometric substitution, we complete the squares in the quadratic \(x^2+3x+2\)

\(x^2+3x+2=\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{4}+2\)

\(=\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}\)

\(\displaystyle\therefore\,I=\int\dfrac{x}{\sqrt{\left(x+\dfrac{3}{2}\right)^2-\dfrac{1}{4}}}dx\)

Put  \(x+\dfrac{3}{2}=t\)

\(\Rightarrow\,dx=dt\)

\(\Rightarrow\,\displaystyle\therefore\,I=\int\dfrac{\left(t-\dfrac{3}{2}\right)dt}{\sqrt{t^2-\dfrac{1}{4}}}\)

Now, put \(t=\dfrac{1}{2}sec\theta\)

and proceed.

  • This method works when you have a quadratic expression inside a radical sign, either in the numerator or in the denominator.

  • Finding integrals by completing the squares and then using trigonometric substitution.
  • Consider an integral.

\(\displaystyle I=\int \dfrac{dx}{\sqrt{6+x-x^2}}\)

Before we make a trigonometric substitution, we complete. The square in the quadratic  \(6+x-x^2\)

\(6+x-x^2=-(x^2-x-6)\)

\(=-\left(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}-6\right)\)

\(=-\left(\left(x-\dfrac{1}{2}\right)^2-\dfrac{25}{4}\right)\)

\(=\dfrac{25}{4}-\left(x-\dfrac{1}{2}\right)^2\)

\(\displaystyle\therefore\,I=\int\dfrac{dx}{\sqrt{\dfrac{25}{4}-\left(x-\dfrac{1}{2}\right)^2}}\)

Now ,we can put \(x-\dfrac{1}{2}=\dfrac{5}{2}sin\theta\)

and then proceed.

Illustration Questions

Evaluate  \(\displaystyle I=\int\dfrac{1}{\sqrt{9+8x-x^2)}}dx\)

A \(sin^{-1}\left(\dfrac{x-4}{5}\right)+C\)

B \(\sqrt{x^2-9x-x^2}+C\)

C \(sin^{-1}\left(\dfrac{x-5}{4}\right)+C\)

D \(2sin^2\,cos^3x+C\)

×

\(\displaystyle I=\int\dfrac{1}{\sqrt{-x^2+8x+9}}dx\)

\(\displaystyle \therefore\,I=\int\dfrac{dx}{\sqrt{-(x^2-8x-9)}}\)

\(\displaystyle I=\int\dfrac{dx}{\sqrt{-\left((x-4)^2-16-9\right)}}\)

\(\displaystyle =\int\dfrac{dx}{\sqrt{-\left((x-4)^2-25\right)}}\)

\(\displaystyle =\int\dfrac{dx}{\sqrt{25-(x-4)^2}}\)

Put  \(x-4=t\)

\(\Rightarrow\,dx=dt\)

\(\displaystyle I=\int\dfrac{dt}{\sqrt{25-t^2}}\)

Put  \(t=5sin\,\theta\)

\(\Rightarrow\,dt=5cos\,\theta\,d\theta\)

\(\displaystyle \therefore\,I=\int\dfrac{5cos\theta\,d\theta}{\sqrt{25-25sin^2\theta}}\)

\(\displaystyle =\int\dfrac{5cos\,\theta\,d\theta}{5cos\,\theta}\)

\(\int d\theta=\theta+C\)

\(\therefore\,I=\theta+C\\=sin^{-1}\dfrac{t}{5} + C\)

\(=sin^{-1}\left(\dfrac{x-4}{5}\right)+C\) (go back to original variable \(x\))

Evaluate  \(\displaystyle I=\int\dfrac{1}{\sqrt{9+8x-x^2)}}dx\)

A

\(sin^{-1}\left(\dfrac{x-4}{5}\right)+C\)

.

B

\(\sqrt{x^2-9x-x^2}+C\)

C

\(sin^{-1}\left(\dfrac{x-5}{4}\right)+C\)

D

\(2sin^2\,cos^3x+C\)

Option A is Correct

Illustration Questions

Evaluate \(\displaystyle I=\int\dfrac{dx}{\sqrt{x^2+4x+6}}\)

A \(xsin^2x+cosx+C\)

B \(x^2e^{2x}-lnx+C\)

C \(ln|(x+6)+\sqrt{x^2+2x+3}|+C\)

D \(ln\left|(x+2)+\sqrt{x^2+4x+6}\,\right|+C\)

×

\(\displaystyle I=\int\dfrac{dx}{\sqrt{x^2+4x+6}}\)

\(\displaystyle I=\int\dfrac{dx}{\sqrt{(x+2)^2-4+6}}\to\) complete the squares 

\(\displaystyle I=\int\dfrac{dx}{\sqrt{(x+2)^2+2}}\)

 

 

Put  \(x+2=t\)

\(\Rightarrow\,dx=dt\)

\(\displaystyle \therefore\,I=\int\dfrac{dt}{\sqrt{t^2+2}}\)

Put  \(t=\sqrt2tan\,\theta\)

\(\Rightarrow\,dt=\sqrt2sec^2\theta\,d\theta\)

\(\displaystyle \therefore\,I=\int\dfrac{\sqrt2sec^2\theta\,d\theta}{\sqrt{2tan^2\theta+2}}\)

\(\displaystyle =\int\dfrac{\sqrt2sec^2\theta\,d\theta}{\sqrt{2sec^2\theta}}\)

\(=\style =\int sec\,\theta\,d\theta\)

\(\therefore\,I=ln|sec\theta+tan\theta|+C\)

\(tan\theta=\dfrac{t}{\sqrt2}\)

\(\Rightarrow\,sec\theta=\sqrt{1+\dfrac{t^2}{2}}\\=\sqrt{\dfrac{t^2+2}{2}}\)

\(\therefore\,I=ln\left|\sqrt{\dfrac{t^2+2}{2}}+\dfrac{t}{\sqrt2}\right|+C\)

\(=ln\left|t+\sqrt{t^2+2}\right|-ln\sqrt2+C\)

\(=ln\left|t+\sqrt{t^2+2}\right|+C\)

\((-ln\sqrt2\,\text{merges with C})\)

\(=ln\left|(x+2)+\sqrt{(x+2)^2+2}\,\right|+C\)   (go back to original variable \(x\))

\(=ln\left|(x+2)+\sqrt{x^2+4x+6}\,\right|+C\)

Evaluate \(\displaystyle I=\int\dfrac{dx}{\sqrt{x^2+4x+6}}\)

A

\(xsin^2x+cosx+C\)

.

B

\(x^2e^{2x}-lnx+C\)

C

\(ln|(x+6)+\sqrt{x^2+2x+3}|+C\)

D

\(ln\left|(x+2)+\sqrt{x^2+4x+6}\,\right|+C\)

Option D is Correct

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