Question
Fill in the blanks:
$\int\limits^{\frac{\pi}{2}}_0\cos\text{x e}^{\sin\text{x}}\text{dx}$ is equal to_________.
$\int\limits^{\frac{\pi}{2}}_0\cos\text{x e}^{\sin\text{x}}\text{dx}$ is equal to_________.
✓
Answer
$\int\limits^{\frac{\pi}{2}}_0\cos\text{x e}^{\sin\text{x}}\text{dx}$ is equal to, $\text{e}-1$
Solution:
Let $\text{I}=\int\limits^{\frac{\pi}{2}}_0\cos\text{x e}^{\sin\text{x}}\text{dx}$
Put $\sin\text{x}=\text{t}\Rightarrow\cos\text{x dx}=\text{dt}$
As $\text{x}\rightarrow0,$ then $\text{t}\rightarrow0$
and $\text{x}\rightarrow\frac{\pi}{2},$ then $\text{t}\rightarrow1$
$\therefore\ \text{I}=\int\limits^1_0\text{e}^\text{t}\text{dt}=\big[\text{e}^\text{t}\big]^1_0$
$=\text{e}^1-\text{e}^0=\text{e}-1$
Solution:
Let $\text{I}=\int\limits^{\frac{\pi}{2}}_0\cos\text{x e}^{\sin\text{x}}\text{dx}$
Put $\sin\text{x}=\text{t}\Rightarrow\cos\text{x dx}=\text{dt}$
As $\text{x}\rightarrow0,$ then $\text{t}\rightarrow0$
and $\text{x}\rightarrow\frac{\pi}{2},$ then $\text{t}\rightarrow1$
$\therefore\ \text{I}=\int\limits^1_0\text{e}^\text{t}\text{dt}=\big[\text{e}^\text{t}\big]^1_0$
$=\text{e}^1-\text{e}^0=\text{e}-1$
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