Question

hey guys more questions on integration

Answer 1

here it is :\[\int\limits dx/1+\cos^2x\]

Answer 2

@rational @iambatman

Answer 3

divide top and bottom \(\cos^2x\)
then try substituting \(\tan x = u\)

Answer 4

\[\int \dfrac{1}{1+\cos^2x}dx = \int \dfrac{\sec^2 x}{\sec^2 x+1}dx = \int \dfrac{\sec^2x}{2+\tan^2 x}dx\]

Answer 5

You could also use cos^2x = 1-sin^2x identity but it will get messy, since you'll have 2- sin^2x, and then you'll have to use another identity

Answer 6

yep
i got it @rational
but wait to c the working

Answer 7

\[\int\limits \sec^2x/\sec^2x+1 (dx)\rightarrow \int\limits \sec^2xdx/2+\tan^2x\]

Answer 8

\[\cos^2x = 1 - \sin^2x\]
\[\large \int\limits \frac{ 1 }{ 2-\sin^2x }dx \implies \int\limits \frac{ 1 }{ 2+\frac{ -1+\cos(2x) }{ 2 } }dx\] alternate method if you're interested, but I like the tanx sub better

Answer 9

\[\let tanx = t\]