The value of displaystyle ∫ (ln (c o t x)/sin â¡ 2 x)dx is equal to (where, C is the constant of integration)

Question

The value of displaystyle ∫ (ln (c o t x)/sin â¡ 2 x)dx is equal to (where, C is the constant of integration) (A) ((ln (cot â¡ x))2/2)+C (B) ((ln (cot â¡ x))2/4)+C (C) ((ln (cot â¡ x))2/6)+C (D) -(1/4)(ln (cot â¡ x))2+C

Tardigrade Admin · Accepted Answer

displaystyle ∫ (ln (cot â¡ x)/2 sin â¡ x cos â¡ x) d x = Î (let) Put ln (cot â¡ x)=t ⇒ (1/cot x)× cosec2â¡xdx=-dt ⇒ (d x/cos x sin â¡ x)=-dt So, Î=(1/2) displaystyle ∫ (- t) d t=(- 1/2)((t2/2))+C =-(1/4)(ln (cot â¡ x))2+C

Q. The value of $\int \frac{l n ( co t x )}{s in 2 x} d x$ is equal to (where, $C$ is the constant of integration)

$\frac{( l n ( co t x ) ) ^{2}}{2} + C$

$\frac{( l n ( co t x ) ) ^{2}}{4} + C$

$\frac{( l n ( co t x ) ) ^{2}}{6} + C$

$- \frac{1}{4} (l n (co t x))^{2} + C$

Q. The value of ∫sin⁡2xln(cotx)​dx is equal to (where, C is the constant of integration)

2(ln(cot⁡x))2​+C

4(ln(cot⁡x))2​+C

6(ln(cot⁡x))2​+C

−41​(ln(cot⁡x))2+C

∫2sin⁡xcos⁡xln(cot⁡x)​dx=I (let) Put ln(cot⁡x)=t ⇒cotx1​×cosec2⁡xdx=−dt ⇒cosxsin⁡xdx​=−dt So, I=21​∫(−t)dt=2−1​(2t2​)+C =−41​(ln(cot⁡x))2+C

Q. The value of $\int \frac{l n ( co t x )}{s in 2 x} d x$ is equal to (where, $C$ is the constant of integration)

$\frac{( l n ( co t x ) ) ^{2}}{2} + C$

$\frac{( l n ( co t x ) ) ^{2}}{4} + C$

$\frac{( l n ( co t x ) ) ^{2}}{6} + C$

$- \frac{1}{4} (l n (co t x))^{2} + C$