Fill In The Blanks[1 Marks ]

Question 11 Mark

Fill in the blanks:
$\int\frac{\text{x}+3}{(\text{x}+4)^2}\text{e}^\text{x}\text{dx}=$ ________.

Answer

$\int\frac{\text{x}+3}{(\text{x}+4)^2}\text{e}^\text{x}\text{dx}=\text{e}^{\text{x}}\frac{1}{\text{x}+4}+\text{C}$
Solution:
$\text{I}=\int\frac{\text{x}+3}{(\text{x}+4)^2}\text{e}^\text{x}\text{dx}$
$=\int\frac{\text{x}+4-1}{(\text{x}+4)^2}\text{e}^\text{x}\text{dx}$
$=\int\text{e}^\text{x}\Big(\frac{1}{\text{x}+4}-\frac{1}{(\text{x}+4)^2}\Big)\text{dx}$
$=\int\text{e}^\text{x}\big(\text{f(x)}+\text{f}'(\text{x})\big)\text{dx,}$ where $=\text{e}^{\text{x}}\frac{1}{\text{x}+4}+\text{C}$

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Question 21 Mark

Fill in the blanks:
$\int\limits^{\frac{\pi}{2}}_0\cos\text{x e}^{\sin\text{x}}\text{dx}$ 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 $\text{t}\rightarrow1$
$\therefore\ \text{I}=\int\limits^1_0\text{e}^\text{t}\text{dt}=\big[\text{e}^\text{t}\big]^1_0$
$=\text{e}^1-\text{e}^0=\text{e}-1$

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Question 31 Mark

Fill in the blanks:
If $\int\limits^\text{a}_0\frac{1}{1+4\text{x}^2}\text{dx}=\frac{\pi}{8},$ then a = ________.

Answer

If $\int\limits^\text{a}_0\frac{1}{1+4\text{x}^2}\text{dx}=\frac{\pi}{8},$ then $\text{a}=\frac{1}{2}$
Let $\text{I}=\int\limits^\text{a}_0\frac{1}{1+4\text{x}^2}\text{dx}=\frac{\pi}{8}$
Now, $\int\limits^\text{a}_0\frac{1}{4\Big(\frac{1}{4}+\text{x}^2\Big)}\text{dx}=\frac{2}{4}\big[\tan^{-1}2\text{x}\big]^\text{a}_0$
$=\frac{1}{2}\tan^{-1}2\text{a}-0=\frac{\pi}{8}$
$\frac{1}{2}\tan^{-1}2\text{a}=\frac{\pi}{8}$
$\Rightarrow\ \tan^{-1}2\text{a}=\frac{\pi}{4}$
$\Rightarrow\ 2\text{a}=1$
$\therefore\ \text{a}=\frac{1}{2}$

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Question 41 Mark

Fill in the blanks:
$\int\frac{\sin\text{x}}{3+4\cos^2\text{x}}\text{dx}=$ ________.

Answer

$\int\frac{\sin\text{x}}{3+4\cos^2\text{x}}\text{dx}=-\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$
Solution:
Let $\text{I}=\int\frac{\sin\text{x}}{3+4\cos^2\text{x}}\text{dx}$
Put $\cos\text{x}=\text{t}\Rightarrow-\sin\text{x dx}=\text{dt}$
$\therefore\ \text{I}=-\int\frac{\text{dt}}{3+4\text{t}^2}=-\frac{1}{4}\int\frac{\text{dt}}{\Big(\frac{\sqrt{3}}{2}\Big)^2+\text{t}^2}$
$=-\frac{1}{4}\cdot\frac{2}{\sqrt{3}}\tan^{-1}\frac{2\text{t}}{\sqrt{3}}+\text{C}=-\frac{1}{2\sqrt{3}}\tan^{-1}\Big(\frac{2\cos\text{x}}{\sqrt{3}}\Big)+\text{C}$

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Question 51 Mark

Fill in the blanks:
The value of $\int\limits^\pi_{-\pi}\sin^3\text{x}\cos^2\text{x dx}$ is _______.

Answer

The value of $\int\limits^\pi_{-\pi}\sin^3\text{x}\cos^2\text{x dx}$ is $0.$
Solution:
Let $\text{I}=\int\limits^\pi_{-\pi}\sin^3\text{x}\cos^2\text{x dx}$
$=\int\limits^\pi_{-\text{x}}\sin^3(\pi-\pi-\text{x})\cos^2(\pi-\pi-\text{x})\text{dx}$ $\bigg[\because\int\limits^\text{b}_\text{a}\text{f(x)}\text{dx}=\int\limits^\text{b}_\text{a}\text{f(a}+\text{b}-\text{x)}\text{dx}\bigg]$
$=\int\limits^\pi_{-\pi}\sin^3(-\text{x})\cos^2(-\text{x})\text{dx}=-\int\limits^\pi_{-\pi}\sin^3\text{x}\cos^2\text{x dx}=-\text{I}$
$\therefore\ 2\text{I}=0$
$\therefore\ \text{I}=0$

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Maths Fill In The Blanks[1 Marks ]

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