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Problem 1: Evaluate $\int e^{2x} \,dx$

Solution:

Let $2x=t$.

$\int \frac{\tan x}{\cot x} dx$

$=\int \tan^2 x \ dx$

$= \int (\sec^2 x-1)\ dx $

 

Problem 2: Show that $\log_3 \log_2 8=1$

Solution:

At first, we calculate $\log_2 8$.

Now, $\log_2 8=\log_2 2^3$

$=3 \log_2 2 \quad$ $[\because \log_a b^k=k \log_a b]$

$=3 \cdot 1 \quad$ $[\because \log_a a=1]$ 

$=3$

$\therefore$ LHS $= \log_3 \log_2 8$

$=\log_3 3=1$ (proved)

 

\[{\dfrac{d}{dx}(x^n)=\dfrac{d}{dx}(f(x))}  {=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}}  {=\lim\limits_{h \to 0}\dfrac{(x+h)^n-x^n}{h}}\]

[Let $z=x+h.$ Then $z \to x$ as $h \to 0$]

\[=\lim\limits_{z \to x} \dfrac{z^n-x^n}{z-x}=nx^{n-1}\]

 

 

Problem 2: Show that $\log_3 \log_2 8=1$
Solution:
At first, we calculate $\log_2 8$.
Now, $\log_2 8=\log_2 2^3$
$=3 \log_2 2 \quad$ $[\because \log_a b^k=k \log_a b]$
$=3 \cdot 1 \quad$ $[\because \log_a a=1]$ 
$=3$
$\therefore$ LHS $= \log_3 \log_2 8$
$=\log_3 3=1$ (proved)
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