Question

The value of integral $\int_{}^{}\frac{(1 - \cos\theta)^{2/7}}{(1 + \cos\theta)^{9/7}}$ dq is –

• A. $\frac { 17 } { 11 }$ $\left( \tan\frac{\theta}{2} \right)^{\frac{11}{7}} + c$
• B. $\frac { 7 } { 11 }$ $\left( \cos\frac{\theta}{2} \right)^{\frac{11}{7}} + c$
• C. $\frac { 7 } { 11 }$ $\left( \sin\frac{\theta}{2} \right)^{\frac{11}{7}} + c$
• D. None of these

Answer 1

Answer: (A)

$\frac { 17 } { 11 }$ $\left( \tan\frac{\theta}{2} \right)^{\frac{11}{7}} + c$

Answer 2

Let I = $\int_{}^{}\frac{(1 - \cos\theta)^{2/7}}{(1 + \cos\theta)^{9/7}}$ dq

= $\int_{}^{}\frac{(2\sin^{2}\theta/2)^{2/7}}{(2\cos^{2}\theta/2)^{9/7}}$dq = $\frac { 1 } { 2 }$ $\int_{}^{}\frac{(\sin\theta/2)^{4/7}}{(\cos\theta/2)^{18/7}}$dq

Put $\frac{\theta}{2}$= t, \ $\frac{d\theta}{2}$ = dt

\ I = $\int_{}^{}\frac{(\sin t)^{4/7}}{(\cos t)^{18/7}}$dt (Here m + n = –2)

$\int_{}^{}{(\tan t)^{4/7}}$sec² tdt

Put tan t = u \ sec²x dt = du

$\int_{}^{}{u^{4/7}du}$= $\frac{u^{11/4}}{11/7}$ + c = $\frac{7}{11}$ (tan t)^11/7 + c

= $\frac { 7 } { 11 }$ $\left( \tan\frac{\theta}{2} \right)^{11/7}$ + c

I_4,3 = $\int_{}^{}{\cos^{4}x}$sin3x dx