Answer to Question #104834 in Calculus for SHIVAM KUMAR

Question

Integrate the following functions w.r.t x:
(A) 1\(2x+1)^3\2
(B)sin (2x+3)
(C)cosec (4x)
(D)1\√1-9x^2
(E)1\1+4x^2

Answer 1

A.) $\int \frac{1}{(2x+1)^{\frac{3}{2}}} \, dx$

$= \int (2x+1)^{- \frac{3}{2}} \, dx$

Let us assume that:

$2x + 1 = u \\ \Rightarrow 2 \, dx = du \\ \Rightarrow dx = \frac{du}{2}$

Now:

Substituting $u = 2x + 1$ into the integration and integrating with respect to $du$ , we have:

$= \frac{1}{2} \int u^{-\frac{3}{2}} \, du$

$= \frac{1}{2} \left[ \frac{u^{-\frac{1}{2}}}{- \frac{1}{2}} \right] + C$ $\hspace{1 cm}$ [ $\because \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \, \,$ ( Where $C$ is a constant of integration) ]

$= - \frac{1}{u^{\frac{1}{2}}} + C \\ = - \frac{1}{\sqrt{u}} + C$

Undo substitution and we have:

$= - \frac{1}{\sqrt{2x+1}} + C \hspace{1 cm}$ [ $\because u = 2x + 1$ ]

$\\$

B.) $\int sin (2x + 3) \, dx$

Let us assume that:

$u = 2x + 3 \\ \Rightarrow du = 2 \, dx \\ \Rightarrow dx = \frac{du}{2}$

Now:

Substituting $u = 2x + 3$ into the integration and integrating with respect to $du$ , we have:

$= \frac{1}{2} \int \sin (u) \, du$

$= - \frac{1}{2} \cos (u) + C \hspace {1 cm}$ [ $\because \int \sin (x) \, dx = - \cos (x) + C \,\,$ (Where $C$ is a constant of integration) ]

Undo substitution and we have:

$= - \frac{\cos(2x+3)}{2} + C \hspace {1 cm} \left[ \because u = 2x+ 3 \right]$

$\\$

C.) $\int \cosec (4x) \, dx$

Let us assume that:

$4x = u \\ \Rightarrow 4 \, dx = du \\ \Rightarrow dx = \frac{du}{4}$

Now:

Substituting $4x = u$ into the integration and integrating with respect to $du$ , we have:

$= \frac{1}{4} \int \cosec (u) \, du$

$= - \frac{1}{4} \ln \mid \cosec (u) + \cot (u) \mid + C \hspace{1 cm}$ [ $\because \int \cosec (x) \, dx = - \ln \mid \cosec (x) + \cot(x) \mid + C \,\,$ (Where $C$ is a constant of integration)]

Undo substitution and we have:

$= - \frac{\ln \mid \cosec (4x) + \cot (4x) \mid }{4} + C \hspace{1 cm} \left[ \because u = 4x \right]$

$\\$

D.) $\int \frac{1}{\sqrt{1 - 9x^2}} \, dx$

$= \int \frac{1}{\sqrt{9(\frac{1}{9} - x^2)}} \, dx$

$= \frac{1}{3} \int \frac{1}{\sqrt{(\frac{1}{3})^2 - x^2}} \, dx$

$= \frac{1}{3} \sin^{-1} \left( \frac{x}{\frac{1}{3}} \right) + C \hspace{1 cm}$ [ $\because \int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \sin^{-1} \left( \frac{x}{a} \right) + C \,\,$ (Where $C$ is a constant of integration)]

$= \frac{1}{3} \sin^{-1} (3x) + C$

$\\$

E.) $\int \frac{1}{1 + 4x^2} \, dx$

$= \frac{1}{4} \int \frac{1}{\frac{1}{4} + x^2} \, dx$

$= \frac{1}{4} \int \frac{1}{(\frac{1}{2})^2 + x^2} \, dx$

$= \frac{1}{4} \left[ \frac{1}{\frac{1}{2}} \tan^{-1} \left( \frac{x}{\frac{1}{2}} \right) \right] + C \hspace{1 cm}$ [ $\because \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \,\,$ (Where $C$ is a constant of integration) ]

$= \frac{\tan^{-1} (2x)}{2} + C$