1. ## integrating sinx(1+cosx)

hey .
I need some help in integrating sinx(1+cosx) I've tried doing integration by parts but got in a big mess.
Help would be greatly appreciated

thankyou
ellie

2. Originally Posted by elliegurl
hey .
I need some help in integrating sinx(1+cosx) I've tried doing integration by parts but got in a big mess.
Help would be greatly appreciated

thankyou
ellie
Have you got some working to show?

I would expand to get $\int \sin x - \cos x \sin x ~dx = \int \sin x ~dx-\int \cos x \sin x ~dx$

u-sub for 2nd part.

3. Originally Posted by elliegurl
$\int{\sin{x}(1+\cos{x})}\;{dx}$
By substitution:
Spoiler:
Let $u = \cos{x}$, then $\dfrac{du}{dx} = -\sin{x} \Rightarrow {dx} = -\dfrac{du}{\sin{x}}.$
So $\int{\sin{x}(1+\cos{x})}\;{dx} = -\int\dfrac{\sin{x}(1+u)}{\sin{x}}\;{du} = -\int{\left(1+u\right)}\;{du} = -\left(u+\dfrac{u^2}{2}\right)+C$

Therefore $\int{\sin{x}(1+\cos{x})}\;{dx} = \boxed{-\dfrac{1}{2}\cos^2{x}-\cos{x}+C}.$

4. Originally Posted by elliegurl
hey .
I need some help in integrating sinx(1+cosx) I've tried doing integration by parts but got in a big mess.
Help would be greatly appreciated

thankyou
ellie
Another way:

$\int \sin(x) + \int \sin(x) \cos(x)$

$\sin(x) \cos(x) = \frac{1}{2} \sin(2x)$

$\int \sin(x) + \frac{1}{2} \int \sin(2x)$

Spoiler:
$-\cos(x) - \frac{1}{4} \cos(2x) + k$

which is equivalent to TheCoffeeMachine's answer. I have used k as the constant of integration because I have shown that the two answers are equivalent below

$= -\cos(x) - \frac{1}{4} (2\cos^2(x) -1) + k$

$= -\frac{1}{2}\cos^2(x) - \cos(x) + \frac{1}{4}+k$

$= -\frac{1}{2}\cos^2(x) - \cos(x) + C$ where $C=\frac{1}{4}+k$

5. ## thankyou

Hey thanks for all your help. the limits were pi and 0 and when i sub them in to both of your answers I get 2 which is right but in the markscheme the integral they have is
-0.5(1+cosx)^2 . I cant see how they have got that. Im a bit confused

6. Originally Posted by elliegurl
Hey thanks for all your help. the limits were pi and 0 and when i sub them in to both of your answers I get 2 which is right but in the markscheme the integral they have is
-0.5(1+cosx)^2 . I cant see how they have got that. Im a bit confused
Put (1+cosx) = t
then -sinx*dx = dt
So the integration becomes
-intg(t)dt. That is equal to
-0.5t^2
= -0.5(1+cosx)^2

7. Originally Posted by sa-ri-ga-ma
Put (1+cosx) = t
then -sinx*dx = dt
So the integration becomes
-intg(t)dt. That is equal to
-0.5t^2
= -0.5(1+cosx)^2

thankyou so much

8. For Solution see the attachment