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Pattern – 8 : Indefinite Integration

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⋆ Pattern – 8

Trigonometric substitution

Let , x = 4 sec θ

x^2=\frac{16}{\frac{3}{2}}

(16 \sec^2{\theta\ } - 16 )

= \left(4^2\right)^\frac{3}{2}\ \left(\sec^2{\theta\ }-1\right)

= 64 \tan^2{\theta\ }

1 - \sin^2{\theta\ } = \cos^2{ \theta\ } const - var

1-\cos^2{\theta\ }=\sin^2{\theta\ }

\sec^2{\theta\ }-1=\tan^2{\theta\ } v - c

cosec^2\ \theta-1=\cot^2{\theta\ }

1 + \tan^2{\theta\ }=\sec^2{\theta\ } c + v = v + c

1 + \cot^2{\theta\ }=cosec^2\ \theta\

\sqrt{x^2+a^2}\rightarrow x = a tan \theta

 

\sqrt{a^2-a^2\sin^2{\theta\ }}

=a\ \sqrt{\ 1-\sin^2{\theta\ }}

=a\ \sqrt{\cos^2{\theta\ }}

=\ \cos{\theta\ }

\sqrt{a^2+a^2+an^2\theta\ }

=a\ \sqrt{\ 1+\tan^2{\theta\ }}

=a\ \sqrt{\sec^2{\theta\ }}

=a\sec{\theta\ }

\sqrt{a^2\sec^2{\theta\ }-a^2}

=a\ \sqrt{\sec^2{\theta\ }-1\ }

=a\tan{\theta\ }

\sqrt{a+a\cos{2\theta\ }}

=\ \sqrt a \ \sqrt{ 1+\cos{2\theta }}

=\ \sqrt{a\ }\ \sqrt{2\cos^2{\theta\ }}

=\ \sqrt{2a}\cos{\theta\ }

 

Some substitutions

1)\ \sqrt{a^2+x^2}\rightarrow\ x\ =\ a\ sin\ \theta\ \ or\ x\ =\ a\ cos\ \theta

2)\ \sqrt{a^2+x^2}\ \rightarrow\ x\ = \ a\ tan\ \theta\ \ or\ x\ =\ a\ cot\ \theta\

3)\ \sqrt{x^2-a^2}\ \rightarrow\ x\ = \ a\ sec\ \theta\ \ or\ x\ =\ a\ cosec\ \theta\

4)\ \sqrt{a+x}\ \ \rightarrow\ x\ = \ a\ {tan}^2{\theta\ }\ or

x=a{cos}^2{\theta\ \ \ or}

x\ =\ cos\ 2\theta\

5)\ \sqrt{a-x}\ \ \rightarrow\ \ x\ =\ a\ {sin}^2{\theta\ }\ or\

x=a{cos}^2 {\theta\ }\ or

x\ =\ a\ cos\ 2\theta

6)\ \sqrt{\frac{a-x}{a+x}}\rightarrow\ \ x\ =\ a\ cos\ 2\theta\

\sqrt{\frac{a+x}{a-x}}\ \ \rightarrow\ \ x\ =\ a\ cos\ 2\theta\

7)\ \sqrt{\frac{x-1}{b-x}}\ \ or\ \ \sqrt{\left(x-a\right)\ \left(x-b\right)}

x\ =\ a\ {cos}^2{\theta\ }+b{sin}^2{\theta\ }

 

Let, x = 3 sin θ

∴  dx = 3 cos θ  d θ

\therefore\ \ I=\ \int\frac{3\cos{\theta\ }}{\left(3\sin{\theta\ }\right)^2}\times3\cos{\theta\ }\ d\ \theta\

=\ \int\frac{9\cos^2{\theta\ }}{9\sin^2{\theta\ }}\ d\ \theta\

∴  I = – cot θ + θ  + c

=\ \frac{-\sqrt{9-x^2}}{x}\ + \ \sin^{-1}{\frac{x}{3}}\ +\ c

\because\frac{x}{3}= \sin{\theta\ }\ \Rightarrow\ \sin^{-1}{\frac{x}{3}}=\ \theta\

\therefore\ \cot{\theta\ } = \ \frac{\sqrt{9-x^2}}{x}

 

 

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