Evaluate the integral with help of trigonometry substitution. \displaystyle \int \frac { x ^ {...

{eq}\Rightarrow \ I= \int \frac{ x ^ { 2 } } { ( x ^ { 2 } - 1 ) ^ { 2 } \sqrt { x ^ { 2 } - 1 } } d x\\ \text{Substitute} \ x=sec(u)\\ \Rightarrow \ u=sec^{-1}x\\ \text{Differentiate with respect to u}\\ \Rightarrow \ \frac{dx}{du}=sec(u)\cdotp\ tan(u)\\ \Rightarrow \ dx=sec(u)\cdotp\ tan(u)du\\ \text{Now, put this value in I}\\ \Rightarrow \ I= \int \frac{sec^{3}(u)\cdotp\ tan(u)}{ (sec^{2}(u)-1 )\sqrt{sec^{2}(u)-1 }}du\\ \text{As we know that}, \\ sec^{2}(u)=sec(u)\cdotp\ tan(u)\\ \Rightarrow \ I= \int \frac{sec^{3}(u)\cdotp\ tan(u)}{tan^{4}(u)\cdotp\ tan(u)}du\\ \Rightarrow \ I= \int \frac{sec^{3}(u)}{tan^{4}(u)}du\\ \Rightarrow \ I= \int \left(\frac{1}{cos^{3}(u)} \right)\cdotp \left(\frac{cos^{4}(u)}{sin^{4}(u)} \right)du\\ \Rightarrow \ I= \int \frac{cos(u)}{sin^{4}(u)}du\\ \text{Now,} \\ \text{Let} \ sin(u)=t\\ \text{Differentiate with respect to t}\\ \Rightarrow \ sin(u)=t\\ \Rightarrow \ cos(u)\frac{du}{dt}=1\\ \Rightarrow \ cos(u)du=dt\\ \text{put this value in I}\\ \Rightarrow \ I= \int \frac{dt}{t^{4}}dt\\ \Rightarrow \ I= \int\ t^{-4}dt\\ {/eq}

{eq}\Rightarrow \ \left[\frac{t^{-4+1}}{-4+1}\ \right]+C\\ \Rightarrow \ \frac{t^{-3}}{-3}+C\\ \Rightarrow \ -\frac{1}{3t^{3}}+C\\ \text{Now put }t=sin(u) \text{ in above solution}\\ \Rightarrow \ -\frac{1}{3sin^{3}(u)}+C\\ \text{Now put} \\ \Rightarrow \ u=sec^{-1}x \ \text{ in the above solution}\\ \Rightarrow \ -\frac{1}{3sin^{3}(sec^{-1}x)}+C\\ \text{As we know that}\\ \Rightarrow \ sec(x)=\frac{\text{ hypotenuse of the right angled triangle}}{\text{base of the right angled triangle}}\\ \Rightarrow \ sec^{-1}\frac{x}{1}\\ \text{Here, hypotenuse } =x \text{ and base}=1\\ \text{then perpendicular}=\sqrt{\text{hypotenuse}^{2}-\text{base}^{2}}\\ \Rightarrow \ \text{perpendicular}=\sqrt{x^{2}-1}\\ \text{then,} \\ \Rightarrow \ sin(x)=\frac{\text{perpendicular of right angled triangle}}{\text{hypotenuse of right angled triangle}}\\ \Rightarrow \ sin(x)=\frac{\sqrt{x^{2}-1}}{x^{2}}\\ \Rightarrow \ sin\left(sec^{-1}(x) \right)=\frac{\sqrt{x^{2}-1}}{x}\\ \Rightarrow \ sin^3 \left(sec^{-1}(x) \right)=\left(\frac{\sqrt{x^{2}-1}}{x} \right)^{3}\\ \Rightarrow \ sin^{3} \left(sec^{-1}(x) \right)=\frac{(x^{2}-1)^\frac{3}{2}}{x^{3}}\\ \text{now put this value in} -\frac{1}{3sin^{3}(u)}+C\\ \Rightarrow \ -\frac{x^{3}}{3\left(x^{2}-1 \right)^\frac{3}{2}}+C\\ {/eq}