this is 0 to pie
this is indefinite
1can u solve both for me please
8 Answers
1can u please type the questions the format of entering the text is hardly readable
11 ∫ log(1 + sinx)dx limit is 0 to pie
2. ∫1 - tan(x +a)tan(x + b)dx
49\int log\left(1+sin\, x \right)dx=log\left(1+sin\,x \right)\int dx-\int \frac{cos\,x}{1+sin\,x}.xdx={log\left(1+sin\,x \right)\int dx-x\int \frac{cos\,x}{1+sin\,x}dx+\int \left\{ \int \frac{cos\,x}{1+sin\,x}dx \right\}dx}
now that can be done! :)
\int \frac{cos\,x}{1+sin\,x}dx = \int \frac{1-six\,x}{cos\,x}dx = \int \left(sec\,x-tan\,x)dx=log\left|sec\,x+tan\,x \right|-log|sec\,x|=log\left|\frac{sec\,x+tan\,x}{sec\,x} \right|=log\left|1+sin\,x \right|
oops!!! ;)
that leads to nowhere! :(
:P :P