Question

Help!

Answer 1

@ganeshie8

Answer 2

screenshot_27
\[\int\limits_{}^{}[\frac{ x^2+\sin^2x }{ 1+x^2 }]\sec^2xdx = \int\limits_{}^{}[\frac{ 1+x^2-1+\sin^2x }{ 1+x^2 }]\sec^2xdx\]

Answer 3

\[\int\limits_{}^{}[\frac{ 1+x^2 }{ 1+x^2 } - \frac{ (1- \sin^2x) }{ 1+x^2 }] \sec^2xdx\]

Answer 4

did u follow the steps

Answer 5

yep,what next?

Answer 6

\[\int\limits_{}^{} [1 -\frac{ \cos^2x }{ 1+x^2 }] \sec^2xdx\]
\[\int\limits_{}^{} \sec^2(x) dx - \frac{ 1 }{ 1+x^2 }dx\]
nowi assume u can do this ques from here

Answer 7

*now i

Answer 8

yes! thanks!

Answer 9

\[\Huge \tan1-\frac{\pi}{4}\]

Answer 10

\[\Large \frac{2}{\cos \theta}=\frac{1}{\cos \theta \cos \phi - \sin \theta \sin \phi}+\frac{1}{\cos \theta \cos \phi + \sin \theta \sin \phi}\]
\[\Large \frac{2}{\cos \theta}=\frac{2\cos \theta \cos \phi}{(\cos \theta \cos \phi)^2- (\sin \theta \sin \phi)^2}\]

Answer 11

here is my attempt for q1)

Answer 12

For the q 25,
\[\Large n=\frac{m!}{(m-2)!2}=>\frac{m(m-1)}{2}\]
and we need..
\[\Large \frac{n!}{(n-2)!2!}=>\frac{n(n-1)}{2}\]

Answer 13

\[\Large (\frac{m(m-1)}{2})\frac{(\frac{m(m-1)}{2}-1)}{2}\]

Answer 14

q1) is done ?

Answer 15

@hartnn

Answer 16

and @ganeshie8 nope stuck there

Answer 17

okay me too.

Answer 18

any idea about determinant one??

Answer 19

yah that one and greatest integer one i got

Answer 20

Since the given integral/function is no where discontinuous within the limits so I guess we can write it as :
\[\Large \int\limits_{0}^{0.9}(x-2x)dx=> \frac{x^2}{2}-x^2]_0^{0.9}\]
where am i wrong? :/

Answer 21

why u ignoring the greatest-integer symbol ?

Answer 22

because [x]=x in the limits 0 to 0.9 :O if it was like 0 to 2 then we could break it from 0 to 1 and 1 to 2 because it wasn't continuous at integral points?

Answer 23

\(\int x \ne \int [x]\)

Answer 24

in the given limits,
as x varies between 0 to 0.9,
[x] stays fixed at 0