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Chapter 8: Problem 31

Find the indefinite integral. $$ \int e^{\sin x} \cos x d x $$

Short Answer

Expert verified
\( \frac{1}{2}e^{\sin x} \sin x + C \)

Step by step solution

01

Identify \(u\) and \(dv\)

Identify two parts of the integral: \(u = e^{\sin x}\) and \(dv = \cos x dx\)
02

Calculate \(du\) and \(v\)

Compute the derivatives and anti-derivatives necessary for the integration by parts formula: \(du = e^{\sin x} \cos x dx\) (found by using the chain rule on \(u\)) and \(v = \sin x\) (which is the anti-derivative of \(\cos x\))
03

Apply integration by parts formula

Recall that the integration by parts formula is given by \(\int udv = uv - \int vdu\). Applying this formula gives the indefinite integral of given function as \( uv - \int v du = e^{\sin x} \sin x - \int \sin x e^{\sin x} \cos x dx = e^{\sin x} \sin x - \int u du \)
04

Simplify the expression

Notice that the integral on the right, \( \int u du \), is the same as our original integral. This allows us to set up an equation and solve for our original integral. Call the original integral \(I\), thus \(I = e^{\sin x} \sin x - I\). Solving this gives \(2I = e^{\sin x} \sin x\).
05

Find the final solution

Isolate \(I\) by dividing through by 2 to find that \(I = \frac{1}{2}e^{\sin x} \sin x + C\), where \(C\) is the constant of integration

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