Answer to Question #176978 in Calculus for Phyroe

Question

Integrals Giving Inverse Trigonometric Functions

∫ dt/√(9t²-16)

Answer 1

We have given the integral,

$I = \int \dfrac{dt}{\sqrt{9t^2-16}}$

$I=\int \dfrac{dt}{\sqrt{-16(1-\dfrac{9t^2}{16})}}$

$I = \int \dfrac{dt}{4i\sqrt{1-(\dfrac{3t}{4})^2}}$

Substituting $x = \dfrac{3t}{4}$

then, $dx = \dfrac{3}{4}dt$

Hence, $I = \dfrac{1}{3i}\int \dfrac{dx}{\sqrt{1-x^2}}$

Now substituting,

$x= sinu\\ dx = cosudu$

then, $I = \dfrac{1}{3i}\int\dfrac{cosu}{cosu}du$

$I = \dfrac{1}{3i}\int du$

$I = \dfrac{1}{3i}[u]+C$

Now putting the value of u in above equation

$I= \dfrac{1}{3i}sin^{-1}x + C$

Now putting the value of x in the above equation,

$I = \dfrac{1}{3i}sin^{-1}(\dfrac{3t}{4})+C$