# Find the following integrals: (1) integral 1 - x/squareroot 1 - x^2 dx. (2) integral sec(2x +...

## Question:

Find the following integrals:

(1) {eq}\displaystyle \int \frac{1 - x}{\sqrt{1 - x^2}}\ dx {/eq}

(2) {eq}\displaystyle \int \frac{\sec(2x + 1)}{\cos^2 (2x + 1)}\ dx {/eq}

(3) {eq}\displaystyle \int 4 \cosh(3t - \ln 2)\ dt {/eq}

(4) {eq}\displaystyle \int \frac{4}{1 + (2x + 1)^2}\ dx {/eq}

(5) {eq}\displaystyle \int^{\displaystyle 0}_{\displaystyle -1}\ \frac{4}{1 + (2x + 1)^2}\ dx {/eq}

## Integration

We will use the substitution method to solve the integrals presented in the questions. We will add a constant C after performing the required integration.

The required steps to evaluate{eq}\int f(x)\quad dx {/eq}by substitution are:

1. Put x=g(y), dx = g'(y) dy in the integrand

2. Evaluate the resulting integral in y.

3. Express the result obtained in terms of x.

Reduction Formula for {eq}\displaystyle{\int \sec^n x dx = \frac{\sec^{n -2}x \tan x}{n -1} + \frac{n-2}{n-1}\int \sec^{n-2} x \ dx\\} {/eq}

(1) {eq}\displaystyle{ \int \frac{1 - x}{\sqrt{1 - x^2}}\ dx\\[10pt] \int \frac{1}{\sqrt{1 - x^2}} \ dx - \int \frac{x}{\sqrt{1 - x^2}} \ dx \hspace{90pt}(1)\\[10pt] I_1 = \int \frac{1}{\sqrt{1 - x^2}} \ dx\\[10pt] I_1 = sin^{-1} x + C\\[10pt] I_2 =\int \frac{x}{\sqrt{1 - x^2}} \ dx\\[10pt] \text{substitute}\quad u = 1 - x^2\\[10pt] \frac{du}{dx} = -2 x\\[10pt] dx = \frac{-1}{2x} \ du\\[10pt] I_2 = \frac{-1}{2}\int \frac{1}{\sqrt{u}} \ du\\[10pt] I_2 = \frac{-1}{2}\int u^\frac{-1}{2} \ du\\[10pt] I_2 = \frac{-1}{2}\left(\frac{2 u^\frac{1}{2}}{1}\right) + C\\[10pt] I_2 = - \sqrt{u} + C\\[10pt] i_2 = - \sqrt{(1 - x^2)} + C \\[10pt] \text{putting the value of}\quad I_1, I_2\quad \text{in (1) we get}\\[10pt] \boxed{sin^{-1} x + \sqrt{(1 - x^2)} + C} \\} {/eq}

(2) {eq}\displaystyle{ \int \frac{\sec(2x + 1)}{\cos^2 (2x + 1)}\ dx \\[10pt] \text{substitute}\quad u = 2x +1\\[10pt] \frac{du}{dx} = 2 \\[10pt] dx = \frac{1}{2} du\\[10pt] I = \frac{1}{2}\int \frac{\sec u}{ \cos ^2 u} \ du\\[10pt] = \frac{1}{2}\int \sec^3 u \ du \hspace{150pt}\left(\because \frac{1}{\cos x} = \sec x\right)\\[10pt] = \frac{1}{2}\left[\frac{\sec u \tan u }{2} + \frac{1}{2}\int \sec u \ dx\right]\hspace{80pt}\left(\int \sec^n x dx = \frac{\sec^{n -2}x \tan x}{n -1} + \frac{n-2}{n-1}\int \sec^{n-2} x \ dx\right)\\[10pt] = \frac{1}{2}\left[\frac{\sec u \tan u }{2} + \frac{1}{2}\left(\ln (\tan u +\sec u)\right)\right] + C\\[10pt] = \frac{1}{2}\left[\frac{\sec(2x + 1) \tan (2x + 1) }{2} + \frac{1}{2}\left(\ln (\tan(2x +1) +\sec (2x +1))\right)\right] + C\\[10pt] = \frac{\sec(2x + 1) \tan (2x + 1) }{4} + \frac{\ln (\tan(2x +1) +\sec (2x +1)}{4} + C\\[10pt] \boxed{= \frac{\sec(2x + 1) \tan (2x + 1) + \ln (\tan(2x +1) +\sec (2x +1)}{4} +C}\\} {/eq}

(3) {eq}\displaystyle{\int 4 \cosh(3t - \ln 2)\ dt\\[10pt] \text{substitute}\quad u = 3t - \ln2\\[10pt] \frac{du}{dt} = 3 -0\\[10pt] dt = \frac{1}{3} \ du\\[10pt] \frac{4}{3} \int cosh(u) \ du\\[10pt] \frac{4}{3} \sinh (u) + C\\[10pt] \boxed{\frac{4 \sinh(3t - \ln2)}{3} + C} \\} {/eq}

(4) {eq}\displaystyle{ \int \frac{4}{1 + (2x + 1)^2}\ dx\\[10pt] \text{substitute}\quad u = 2x + 1\\[10pt] \frac{du}{dx} = 2\\[10pt] dx = \frac{1}{2} \ du\\[10pt] \int \frac{4}{2(1 + u^2)} \ du\\[10pt] 2\int \frac{1}{1 + u^2} \ du\\[10pt] 2 \tan^{-1} u + C\\[10pt] \boxed{2 \tan^{-1}(2x + 1) + C} \\} {/eq}

(5) {eq}\displaystyle{ \int^{ 0}_{ -1}\ \frac{4}{1 + (2x + 1)^2}\ dx\\ \text{substitute}\quad u = 2x + 1\\[10pt] \frac{du}{dx} = 2\\[10pt] dx = \frac{1}{2} \ du\\[10pt] \int^{0}_{-1} \frac{4}{2(1 + u^2)} \ du\\[10pt] 2\int^{0}_{-1} \frac{1}{1 + u^2} \ du\\[10pt] \left[2 \tan^{-1} u \right]_{-1}^{0}\\[10pt]\\ \left[2 \tan^{-1} (2x +1) \right]_{-1}^{0}\\[10pt]\\ \left[2 \tan^{-1} (0 + 1) \right] - \left[2 \tan^{-1} (-2 +1) \right]\\[10pt] \left[2 \tan^{-1} ( 1) \right] - \left[2 \tan^{-1} (-1) \right]\\[10pt] \left[2\left( \frac{\pi}{4}\right) \right] - \left[2\left( \frac{-\pi}{4}\right) \right] \\[10pt] \frac{\pi}{2} + \frac{\pi}{2}\\[10pt] \boxed{\pi}\\} {/eq}