Question

explain clearly.

∫(7𝑐𝑠𝑐2𝑥 + 2 sec 𝑥 tan 𝑥)𝑑𝑥

Answer 1

$$\int(7\csc2x+2 \sec x \tan x)dx$$

$$=\int(\dfrac{7}{2\sin x\cos x}+\dfrac{2\sin x}{\cos^2x})dx$$

$$=\int(\dfrac{7\sin x}{2\sin^2 x\cos x}+\dfrac{2\sin x}{\cos^2x})dx$$

Use $u$ -substitution

$$u=\cos x, du=-\sin x dx$$

$$\int(\dfrac{7\sin x}{2\sin^2 x\cos x}+\dfrac{2\sin x}{\cos^2x})dx$$

$$=\int(-\dfrac{7}{2u(1-u^2)}-\dfrac{2}{u^2})du$$

$$-\dfrac{7}{2u(1-u^2)}=\dfrac{A}{u}+\dfrac{B}{1-u}+\dfrac{C}{1+u}$$

$$=\dfrac{A(1-u^2)+Bu(1+u)+Cu(1-u)}{u(1-u^2)}$$

$$-\dfrac{7}{2}=A(1-u^2)+Bu(1+u)+Cu(1-u)$$

$$u=0:A=-\dfrac{7}{2}$$

$$u=-1:C=\dfrac{7}{4}$$

$$u=1:B=-\dfrac{7}{4}$$

$$\int(-\dfrac{7}{2u(1-u^2)})du$$

$$=-\dfrac{7}{2}\int\dfrac{du}{u}-\dfrac{7}{4}\int\dfrac{du}{1-u}+\dfrac{7}{4}\int\dfrac{du}{1+u}$$

$$=-\dfrac{7}{2}\ln |u|+\dfrac{7}{4}\ln|1-u|+\dfrac{7}{4}\ln|1+u|+C_1$$

$$\int(-\dfrac{2}{u^2})du=\dfrac{2}{u}+C_2$$

Then

$$\int(-\dfrac{7}{2u(1-u^2)}-\dfrac{2}{u^2})du$$

$$=-\dfrac{7}{2}\ln |u|+\dfrac{7}{4}\ln|1-u|+\dfrac{7}{4}\ln|1+u|+\dfrac{2}{u}+C$$

So

$$\int(7\csc2x+2 \sec x \tan x)dx$$

$$=-\dfrac{7}{2}\ln |\cos x|+\dfrac{7}{4}\ln|1-\cos x|+\dfrac{7}{4}\ln|1+\cos x|$$

$$+\dfrac{2}{\cos x}+C$$

$$=-\dfrac{7}{2}\ln |\cos x|+\dfrac{7}{4}\ln|1-\cos^2 x|+\dfrac{2}{\cos x}+C$$

$$=-\dfrac{7}{2}\ln |\cos x|+\dfrac{7}{2}\ln |\sin x|+\dfrac{2}{\cos x}+C$$