Question

I have some problem in integration

Answer 1

Post the problem.

Answer 2

evaluate \[Evaluate : \int\limits_{}^{} e^x \sin(x/2) dx\]

Answer 3

\[\int\limits_{0}^{\pi/6} \cos^3(2x) dx\]

Answer 4

Hint :

first one : integration by parts

second one :\[\cos^3 x=\cos x(1-\sin^2 x)\]

Answer 5

1. \[\int\limits g*H = G*H-\int\limits G*h\] --> -2/5 e^x (cos(x/2)-2 sin(x/2))

Answer 6

is this right for the first one( by part), I =\[\int\limits_{}^{} e^x \sin (x/2) dx = 1/2 e^x \sin (x/2) - 1/2 e^x \cos(x/2) + 1/2c\]

Answer 7

check it again :) must be look like what carl got

Answer 8

I don not understand the answer form carl. could you explain please.

Answer 9

\[I=\int e^x \sin(x/2) \ \text{d}x\]let e^x dx =dv and sin(x/2)=u\[I=\int e^x \sin(x/2) \ \text{d}x=e^x \sin(x/2)-\frac{1}{2} \int e^x \cos(x/2) \ \text{d}x \ \ \ (\star)\]now let\[II=\int e^x \cos(x/2) \ \text{d}x\]again we let e^x dx =dv and cos(x/2)=u\[II=\int e^x \cos(x/2) \ \text{d}x=e^x \cos(x/2)+\frac{1}{2} \int e^x \sin(x/2) \ \text{d}x\]\[II=e^x \cos(x/2)+\frac{1}{2} I\]put this in \((\star)\)\[I=\int e^x \sin(x/2) \ \text{d}x=e^x \sin(x/2)-\frac{1}{2} (e^x \cos(x/2)+\frac{1}{2} I)\]\[\frac{5}{4}I=e^x \sin(x/2)-\frac{1}{2} e^x \cos(x/2)\]

Answer 10

both functions have infinte derivates so the principal for finding the solution is two have two integrals that look the same so you can sum them up and divide the equation by two as shown by mukushla almost correctly...

Answer 11

you are a great people.

Answer 12

edyes, it is as same as my answer but I forgot the 1/2, guys, I am really thankful for you. you are the nation of MAMTH

Answer 13

yes*

Answer 14

Hi guys for this \[\int\limits_{0}^{\pi/6} \cos^3(2x) dx \]

Answer 15

my final answer is \[\sin2x + 1/3 \sin ^3 (2x) + c

Answer 16

is it correct??

Answer 17

If you are finding a definite integral, wouldn't the answer be numerical?
I got [(3 times square root of 3)] / 16.