Integrals Help?

Question

integral of (sec x)^3 dx

integral of (e^x) cos dx

i got 9e^x cos x + e^x sin x)/2 if anyone could tell me that's rite that'd be great. not sure wut to do for the first one

oops, in my answer to the integral i put a 9 instead of a parenthesis

Answer 1

For the first one

sec(x)^3=sec(x)*sec(x)^2.

Using the identity sec(x)^2=1+tan(x)^2 we write

=sec(x)*(1+tan(x)^2)=
sec(x)+sec(x)tan(x)^2.

Now we can break up our integral into 2. Integrating sec(x)(tan(x))^2 isn't so bad if you use u-sub. Integrating sec(x) is a challenge if you haven't seen the derivation before. I'll challenge you to play with that one or look it up in a table of intregrals.

Answer 2

∫sec^3 (x) dx

applying reduction formula

∫sec^n x dx = sec^(n-2) x tanx/(n-1) + n-2/n-1∫ sec^(n-2) x dx

∫sec^3 x dx = sec x tanx /2 + 1/2∫ sec x dx

(1/2) sec x tanx + 1/2( log(secx + tanx) + c

2)
∫ e^x cos x dx

integration by parts

u = cosx ; du = - sinx dx

dv = e^x ; v = e^x

∫ e^x cos x dx = e^x cosx + ∫e^x sin x dx

again integrating by parts

let u = sin x : du = cos x

dv = e^x; v = e^x

∫ e^x cos x dx = e^x cosx + e^x sin x - ∫e^x cos x

2∫ e^x cos x dx = e^x cosx + e^x sin x

= e^x( sin x + cos x)

∫ e^x cos x dx = (e^x/2)( sin x + cos x) + c

Answer 3

The first one is a pain in the αss. I got
(-log(cos(x/2) - sin(x/2)) + log(cos(x/2) + sin(x/2)) + sec(x)tan(x))/2
I think you got the 2'nd one right ☺

Doug

Answer 4

to verify your answer you only have to differentiate.