Question

$\int \tan (ax+b^2+d^2)dx,a\ne 0$ is equal to
(where $C$ is the constant of integration and $a,b$ and $d$ are fixed constants)

• A. $\ln |\sec (ax+b^2+d^2)|+C$
• B. $-\ln |\cos (ax+b^2+d^2)|+C$
• C. $\frac{\ln |\sec (ax+b^2+d^2)|}{a}+C$
• D. $\frac{-\ln |\cos (ax+b^2+d^2)|}{a}+C$

Answer 1

Answer: (C), (D)

$\frac{-\ln |\cos (ax+b^2+d^2)|}{a}+C$

$\int \tan (ax+b^2+d^2)dx=\frac{\ln |\sec (ax+b^2+d^2)|}{a}+C$
$(\because \int \tan (\alpha x+\beta )dx=\frac{\ln |\sec (\alpha x+\beta )|}{\alpha }+C,\alpha \ne 0)$
$\Rightarrow \frac{\ln |\sec (ax+b^2+d^2)|}{a}+C=\frac{\ln \left(\frac{1}{|\cos (ax+b^2+d^2)|}\right)}{a}+C$
$\Rightarrow \frac{-\ln |\cos (ax+b^2+d^2)|}{a}+C$