Question

@phi

Answer 1

bro can you please check my answer for roman 4 the deduce part

Answer 2

?

Answer 3

@phi

Answer 4

\[\int\limits \frac{sinx}{\cos^3x}dx=\int\limits \frac{sinx}{cosx}.\frac{1}{\cos^2x}dx=\int\limits tanx.\sec^2x.dx\]

Answer 5

try solving it now :) much easier this way

Answer 6

@Nishant_Garg so i am right?

Answer 7

I haven't checked, but it's a long way the way you r doing this problem, look at the integral above, if you take tanx=u you can solve it easily in 1 step

Answer 8

@Nishant_Garg yours is simpler:)) but its good that both gave the same answer:))) thanks bro