Learn with the example of using trig substitution to evaluate an indefinite integration. Practice integration by completing the squares and calculus trigonometric substitution.
Some standard substitutions are
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.
A \(x=3sec\,\theta\)
B \(x=tan\theta\)
C \(x=3\,sin\theta\)
D \(x=cot\theta\)
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.
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 \(x\)
\(x=tan\,\theta\)
\(sin\theta=\dfrac{x}{\sqrt{1+x^2}}\)
\(\therefore\,I=-\dfrac{1(1+x^2)^{3/2}}{3x^3}\)
\(=-\dfrac{1}{3}\dfrac{(1+x^2)^{3/2}}{x^3}+C\)
Option A is Correct
\(\displaystyle I=\int\dfrac{x}{\sqrt{x^2+3x+2}}dx\)
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.
\(\displaystyle I=\int \dfrac{dx}{\sqrt{6+x-x^2}}\)
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.
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 \(2sin^2\,cos^3x+C\)
A \(xsin^2x+cosx+C\)
B \(x^2e^{2x}-lnx+C\)
C \(ln|(x+6)+\sqrt{x^2+2x+3}|+C\)
D \(ln\left|(x+2)+\sqrt{x^2+4x+6}\,\right|+C\)
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.
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 \(x\).
\(x=3\,sec\,\theta\)
\(sin\,\theta=\dfrac{\sqrt{x^2-9}}{x}\)
\(\therefore \,I=\dfrac{\sqrt{x^2-9}}{9x}+C\)
Option A is Correct
A \(\dfrac{1}{2}\)
B \(\dfrac{\pi}{16}\)
C \(\dfrac{\pi}{8}\)
D \(\dfrac{3\pi}{4}\)
