Question

1. (51) ∫(cos x)e^{sin x}dx
2. (51) ∫2 tan x sec²x dx

Explanation:

Step1: Solve $\int (\cos x)e^{\sin x}dx$

Let \(u = \sin x\), then \(du=\cos xdx\).
Substitute into the integral: \(\int e^{u}du\).
Since \(\int e^{u}du=e^{u}+C\), substituting back \(u = \sin x\), we get \(e^{\sin x}+C\).

Step2: Solve $\int 2\tan x\sec^{2}x dx$

Let \(u=\tan x\), then \(du = \sec^{2}x dx\).
The integral becomes \(\int 2u du\).
Using the power - rule \(\int ax^{n}dx=\frac{a}{n + 1}x^{n+1}+C\) (\(n
eq - 1\)), for \(a = 2\) and \(n = 1\), \(\int 2u du=2\times\frac{u^{2}}{2}+C=u^{2}+C\).
Substituting back \(u=\tan x\), we get \(\tan^{2}x + C\).

Answer:

For \(\int (\cos x)e^{\sin x}dx\), the answer is \(e^{\sin x}+C\).
For \(\int 2\tan x\sec^{2}x dx\), the answer is \(\tan^{2}x + C\).