Question

Evaluate the Integral \(\int\limits_{\sqrt 2 }^2 {\frac{1}{{{t^3}\sqrt {{t^2} - 1} }}} 'dt\)

Short answer

Evaluate the integral

After evaluating, \(\int\limits_{\sqrt 2 }^2 {\frac{1}{{{t^3}\sqrt {{t^2} - 1} }}} 'dt\)will be equal to \(\frac{\pi }{{24}} + \frac{{\sqrt 3 }}{s} - \frac{1}{4}\)

Step-by-step solution

Step 1:-(substitute t= sec x and then solve)

given:- I=\(\int\limits_{\sqrt 2 }^2 {\frac{1}{{{t^3}\sqrt {{t^2} - 1} }}} 'dt\)

Let (t)= \(\frac{d}{{dt}}\left( {\sec x} \right)\)

Or dt=sec x’ tan x’ dx

So , for t =\(\sqrt {2,,} x = \frac{\pi }{4}\)

For t=2, \(x = \frac{\pi }{3}\)

I=\(\int\limits_{\frac{\pi }{4}}^3 {\frac{{\sec x\tan x'dx}}{{{{\sec }^3}\sqrt {{{\sec }^2}x - 1} }}} \)

\(\left( {as\sqrt {{{\sec }^2}x - 1} = \tan x} \right)\)

Evaluation

I=\(\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {{{\cos }^2}x.dx} \)

I=\(\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{1}{2}} \left( {1 + \cos x} \right).dx\)

I=\(\frac{1}{2}\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {dx + } \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {{{\cos }^2}x.du} \)

I=\(\frac{1}{2}|x|_{\frac{\pi }{4}}^{\frac{\pi }{3}} + \frac{1}{2}|\frac{{\sin 2x}}{e}|_{\frac{\pi }{4}}^{\frac{\pi }{3}}\)

I= \(\frac{1}{2}|\frac{\pi }{3} - \frac{\pi }{4}| + \frac{1}{2}|\frac{{\sin 2\frac{\pi }{3} - \sin \frac{\pi }{2}}}{2}\)

I=\(\frac{\pi }{{24}} + \frac{{\sqrt 3 }}{8} - \frac{1}{4}\)

Answer:-

Hence ,after evaluating , \(\int\limits_{\sqrt 2 }^2 {\frac{{dt}}{{{t^3}\sqrt {{t^2} - 1} }}} \)will be equal to \(\frac{\pi }{{24}} + \frac{{\sqrt 3 }}{8} - \frac{1}{4}\)