Question

If $\int_{}^{}\frac{\cos^{4}x + 1}{\cot x - \tan x}$ dx = a cos²2x + b cos 2x + c log | cos 2x | + l, l being a constant of integration then –

• A. a = –$\frac{1}{8}$
• B. b = –$\frac{1}{8}$
• C. c = $\frac{5}{8}$
• D. a + b + c = 0

Answer 1

Answer: (B)

b = –$\frac{1}{8}$

Answer 2

$\int \frac { \cos ^ { 4 } x + 1 } { \frac { \cos x } { \sin x } - \frac { \sin x } { \cos x } }$dx = $\frac { 1 } { 2 }$ $\int_{}^{}\frac{(\cos^{4}x + 1)\sin 2x}{\cos 2x}$dx

= $\frac { 1 } { 2 }$ $\int_{}^{}\frac{\left\{ \left( \frac{1 + \cos 2x}{2} \right)^{2} + 1 \right\}}{\cos 2x}$sin 2x dx

= $\frac { 1 } { 8 }$ $\int_{}^{}\left\{ \frac{(1 + \cos 2x)^{2} + 4}{\cos 2x} \right\}$sin 2x dx

put cos 2x = t ̃ (– sin 2x) 2dx = dt

sin 2x dx = –$\frac{dt}{2}$

= –$\frac { 1 } { 16 }$ $\int_{}^{}\frac{(1 + t)^{2} + 4}{t}$dt

=–$\frac { 1 } { 16 }$ $\int_{}^{}\left( \frac{1 + t^{2} + 2t + 4}{t} \right)$dt

= –$\frac { 1 } { 16 }$ $\int_{}^{}\left( 2 + t + \frac{5}{t} \right)$dt = –$\frac { 1 } { 16 }$ $\left\lbrack 2t + \frac{t^{2}}{2} + 5\log|t| \right\rbrack$+ l

= –$\frac { 1 } { 32 }$cos² 2x –$\frac{1}{8}$cos 2x –$\frac{5}{16}$log |cos 2x| + l