Question

evaluate the integral: ∫sin⁡2x/√(cos^2 x+16) dx

Answer 1

\[\int\limits \frac{\sin(2x)}{\cos^2x+16}dx\]

Answer 2

Hmm try subbing
\[u=\cos^2x+16\]
\[du=-2\sin(x)\cos(x)dx\]
\[-du=2\sin(x)\cos(x)dx\]
From
\[\int\limits \frac{\sin(2x)}{\cos^2x+16}dx\]
\[=\int\limits \frac{2\sin(x)\cos(x)}{\cos^2x+16}dx\]
Substituting gives
\[=-\int\limits \frac{1}{u} \, du\]

Answer 3

there's root in the denom. i think

Answer 4

Like this
\[\int\limits \frac{ \sin (2 x)}{\sqrt{ \cos ^2x+16}} \, dx\]

Answer 5

yep..

Answer 6

ok...so sin 2x= 2sinx cosx..

Answer 7

let cos x be t...-sin xdx=dt..

Answer 8

then u'll find the expression reducing to t/(t^2+16)^1/2..

Answer 9

now 2tdt=d(t^2)..

Answer 10

u'll then get d(k)/(k+16)^1/2...k=t^2...

Answer 11

You don't have to do that, take the whole thing from the denominator,
\[u=\cos^2x+16\]
\[-du=2\cos(x)\sin(x)dx\]
Which gives
\[\int\limits\frac{-du}{\sqrt u}\]

Answer 12

I didn't mention the constants to be multiplied or divided

Answer 13

lol yeah

Answer 14

Yeah so just integrate and sub back u, done