Question

Calculate the integral of the following functions:
a) sin^2 (x)
b) 1/(√x^2 -1)

Answer 1

Answer on Question #52765 – Math – Integral Calculus

Question

Calculate the integral of the following functions:

a) $\sin^2(x)$

b) $1 / (\sqrt{x^2 - 1})$

Solution

a) $\int \sin^2(x) dx = \int \frac{1 - \cos(2x)}{2} dx = \frac{x}{2} - \frac{1}{4} \cdot \sin(2x) + C = \frac{1}{2} \cdot (x - \frac{\sin(2x)}{2}) + C = \frac{1}{2} \cdot (x - \sin(x)\cos(x)) + C,$

where $C$ is an arbitrary real constant;

b) $\int \frac{1}{(\sqrt{x^2 - 1})} dx = \int \frac{dx}{|x| - 1} = \begin{cases} \ln |x - 1| + C, & \text{if } x > 0 \\ -\ln |x + 1| + C, & \text{if } x < 0 \end{cases}$

To calculate integral of $1 / \sqrt{x^2 - 1}$,

substitute $x = \cosh t$,

$$dx = (\cosh t)_t^2 dt = (\sinh t) dt,$$

apply $\cosh^2 t - \sinh^2 t = 1$,

$$\cosh^2 t - 1 = \sinh^2 t,$$ $$x^2 - 1 = \sinh^2 t,$$ $$\frac{dx}{\sqrt{x^2 - 1}} = \frac{(\sinh t) dt}{\sqrt{\cosh^2 t - 1}} = \frac{\sinh t dt}{\sqrt{\sinh^2 t}} = dt$$

Rewrite

$$x = \cosh t,$$ $$x = \frac{e^t + e^{-t}}{2},$$ $$e^t + \frac{1}{e^t} = 2x,$$ $$e^{2t} - (2x)e^t + 1 = 0,$$ $$e^t = \frac{2x + \sqrt{4x^2 - 4}}{2} = x + \sqrt{x^2 - 1},$$ $$t = \ln |x + \sqrt{x^2 - 1} |.$$

Finally

$$\int \frac{dx}{\sqrt{x^2 - 1}} = \int dt = t + C = \ln |x + \sqrt{x^2 - 1}| + C,$$

where $C$ is an arbitrary real constant.

Answer:

a) $\frac{1}{2} \cdot (x - \sin(x)\cos(x)) + C$

b) $\log(x - 1) + C$

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