How to Evaluate $ \int e^{\sin\theta} \cos\theta \ d\theta \ $

Question

I am new to integration, I want to evaluate

$$ \int e^{\sin\theta} \cos\theta \ d\theta \ $$

I didn't know much methods, such as substitution, etc. So I want a simple way.

Edit: What I done:

Since, $ e^{sin\theta} $ is a number raised to exponent, I wrote

$$ e^{\sin\theta} \int \cos\theta \ d\theta \ $$

Which gives,

$$ e^{\sin\theta} sin\theta + C \ $$ Where C is the constant of integration.

(I also need to know why I was wrong)

Answer 1

Well, since you do not know much about integration techiques, as you stated, you can see, from the chain rule, that: $$(e^{\sin\theta})'=e^{\sin\theta}(\sin\theta)'=e^{\sin\theta}\cos\theta$$ So, it is immediate that: $$\int e^{\sin\theta}\cos\theta d\theta=\int(e^{\sin\theta})'d\theta=e^{\sin\theta}+c\mbox{, $c\in\mathbb{R}$}$$

Otherwise, you can make the substitution: $$u=\sin\theta$$ So $$du=\frac{d}{d\theta}(\sin\theta)d\theta=\cos\theta d\theta$$ And, finally: $$\int e^{\sin\theta}\underbrace{\cos\theta d\theta}_{du}=\int e^udu=e^u+c\overset{u=\sin\theta}{=}e^{\sin\theta}+c,\ c\in\mathbb{R}$$

Hope this helps! :)

Answer 2

If you don't know the classical rules, you can work by trial and error.

As the derivative of an exponential is an exponential, start by studying

$$\left(e^{\sin \theta}\right)'.$$

Using the chain rule, it turns out that

$$\left(e^{\sin \theta}\right)'=\cos\theta\ e^{\sin \theta}$$

which is precisely your integrand.

Can you conclude ?

Answer 3

Assume that $\sin\theta=p$ then $dp=\cos\theta \cdot d\theta$ So your integral $\int e^{\sin\theta}\cdot \cos\theta\cdot d\theta$ is equivalent to $\int e^p.dp$

Answer 4

let $u=\sin(\theta)$ then $du=\cos(\theta)d\theta$

$\implies \int e^{\sin\theta}\cos(\theta)d\theta$

$\implies\int e^udu$

$\implies e^u+C$