Question

Some fun integrals

Answer 1

Here are some fun integrals I encourage everyone to try them
\[\int\limits [\log(logx)+\frac{1}{(logx)^2}]dx\]
and
\[\int\limits [\sqrt{cotx}+\sqrt{tanx}]dx\]

Answer 2

@mathmath333

Answer 3

try to apply integration by parts to this

\(\log(\log x)\)

Answer 4

first one looks hard but interesting :)
here is how im trying the second :

\[\begin{align}
\int (\sqrt{\cot x}+ \sqrt{\tan x})dx &=\int \dfrac{1+\tan x }{\sqrt{\tan x}}dx\\~\\
\end{align}\]

next u sbstitution i guess
\(u^2 = \tan x \implies dx = \dfrac{2u}{1+u^4}du \)
\[\begin{align}
\int \dfrac{1+\tan x }{\sqrt{\tan x}}dx&= \int \dfrac{1+u^2}{u}\dfrac{2u}{1+u^4}du\\~\\
&=2\int \dfrac{1+u^2}{1+u^4}du\\~\\
&=\cdots
\end{align}\]

Answer 5

doing pretty good so far ganeshie, keep going

Answer 6

so nishant garg u have solution for first or asking for help

Answer 7

Ive done these, thought they are good so share with others :)

Answer 8

cool

Answer 9

they are indeed interesting :) let me see if i can complete the second one

\[\begin{align}
\int \dfrac{1+\tan x }{\sqrt{\tan x}}dx&= \int \dfrac{1+u^2}{u}\dfrac{2u}{1+u^4}du\\~\\
&=2\int \dfrac{1+u^2}{1+u^4}du\\~\\
&=2\int \dfrac{\frac{1}{u^2}+1}{\frac{1}{u^2}+u^2}du\\~\\
&=2\int \dfrac{\frac{1}{u^2}+1}{\left(u-\frac{1}{u}\right)^2+2}du\\~\\
&=2\int \dfrac{1}{t^2+2}dt~~\color{gray}{(t =u-\frac{1}{u} )}\\~\\
&=\dfrac{2}{\sqrt{2}} \arctan\left(\frac{t}{\sqrt{2}}\right)+C\\~\\
&=\sqrt{2} \arctan\left(\frac{u-\frac{1}{u}}{\sqrt{2}}\right)+C\\~\\
&=\sqrt{2} \arctan\left(\frac{\tan x-1}{\sqrt{2\tan x}}\right)+C\\~\\
\end{align}\]

Answer 10

yeah, that's right @ganeshie8

Answer 11

put lnx= t dx=e^t.. add subtract 1/t inside the bracket...two integrals will form of type e^x(f(x)+f'(x)) which on integration gives e^xf(x) or simply use by part in one integral of both brackets

Answer 12

here is the solution:
by parts we have:
\[\int {\log \left( {\log x} \right)dx = x\log \left( {\log x} \right) - \int {\frac{{dx}}{{\log x}}} } \qquad (1)\]
and, by integrating byparts we have again:
\[\int {\frac{{dx}}{{\log x}}} = \frac{x}{{\log x}} + \int {\frac{{dx}}{{{{\left( {\log x} \right)}^2}}}} \]
from which, we get:
\[\int {\frac{{dx}}{{{{\left( {\log x} \right)}^2}}}} = \int {\frac{{dx}}{{\log x}}} - \frac{x}{{\log x}}\quad \left( 2 \right)\]
now adding (1)+(2), we get:
\[\int {\left[ {\log \left( {\log x} \right) + \frac{1}{{{{\left( {\log x} \right)}^2}}}} \right]} \;dx = x\log \left( {\log x} \right) - \frac{x}{{\log x}} + \lambda \]

Answer 13

where, as usual, \[\lambda \], is the arbitrary real constant

Answer 14

Good job guys, you nailed it. Haha