CBSE Class 12-science Answered

Evaluate $integral fraction numerator d x over denominator sin x left parenthesis 5 minus 4 cos x right parenthesis end fraction$ Please also tell whether we can evaluate $integral subscript 0 superscript pi fraction numerator x d x over denominator 1 plus e sin x end fraction$ . if this cannot be evaluated , please explain the reason for my understanding . Thanks

Asked by ahuja8087 | 10 Mar, 2017, 06:24: PM

Expert Answer

Dear student, this is the solution to your query.

$Let I = \int \frac{dx}{sinx (5 - 4 cosx)} Using \sin 2 x = 2 sinxcosx and \cos 2 x = \cos^{2} x - \sin^{2} x \Rightarrow I = \int \frac{dx}{2 \sin \frac{x}{2} \cos \frac{x}{2} (5 - 4 (\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}))} Divide by numerator and denominator by \cos^{2} \frac{x}{2} \Rightarrow I = \int \frac{\frac{1}{\cos^{2} \frac{x}{2}}}{2 \tan \frac{x}{2} (5 - 4 (\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}))} dx \Rightarrow I = \int \frac{\sec^{4} \frac{x}{2}}{2 \tan \frac{x}{2} (5 (1 + \tan^{2} \frac{x}{2}) - 4 (1 - \tan^{2} \frac{x}{2}))} dx \Rightarrow I = \int \frac{\sec^{2} \frac{x}{2} . \sec^{2} \frac{x}{2}}{2 \tan \frac{x}{2} (5 + 5 \tan^{2} \frac{x}{2} - 4 + 4 \tan^{2} \frac{x}{2})} dx \Rightarrow I = \int \frac{\sec^{2} \frac{x}{2} . (1 + \tan^{2} \frac{x}{2})}{2 \tan \frac{x}{2} (1 + 9 \tan^{2} \frac{x}{2})} dx Put \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^{2} \frac{x}{2} dx = dt \Rightarrow I = \int \frac{2 (1 + t^{2})}{2 t (1 + 9 t^{2})} dt \Rightarrow I = \int \frac{1 + t^{2}}{t (1 + 9 t^{2})} dt \Rightarrow I = \int \frac{1 + 9 t^{2} - 8 t^{2}}{t (1 + 9 t^{2})} dt \Rightarrow I = \int \frac{1}{t} - \frac{8 t}{(1 + 9 t^{2})} dt \Rightarrow I = \ln | t | - \frac{4}{9} \ln | 1 + 9 t^{2} | \Rightarrow I = \ln | \tan \frac{x}{2} | - \frac{4}{9} \ln | 1 + 9 \tan^{2} \frac{x}{2} | + C$

As for the pending query, we will have our expert who will solve it on tuesday or wednesday and send it to you with your other questions that you post. We will surely update you on that student. As of now, we will have to keep that query pending. Thank you student.

Answered by | 10 Mar, 2017, 09:51: PM