Question

Integration of trig function with a large even exponent. Problem in comment:

Answer 1

\[\int \frac{(cosec^2(x)-2014)}{cos^{2014}(x)}dx\]

Answer 2

sec?

Answer 3

oh csc?

Answer 4

I'm not used to the above abbreviation for cosecant
sorry

Answer 5

Even with answer choices this one is not easy to figure out.
\[A) \Large -sec^{2015}(x)cot(x) + sin(x) + C\\
B) \Large -sec^{2014}(x)cot(x) + C\\
C) \Large -cosec^{2014}(x)tan(x) + C\\
D) \Large -sec^{2014}(x)cosec^2(x) + C\]

Answer 6

Well you could find the derivative of your answer choices to see which one gives you the expression in the integral.

Answer 7

do you really need it now? i'm too sleepy now will check it tomorrow.

Answer 8

let me see if i can actually integrate this sucker

Answer 9

First the derivatives have to be taken and then they have to be rearranged to see if they can be expressed in the form of the original expression. What puzzles me is that the constant 2014 does not appear as a factor in any of the answers.

Answer 10

i think the sub tan(x/2)=u might really help

Answer 11

if i didn't make an error, it seems to be working out very pretty like

Answer 12

And you are getting one of the answers?

Answer 13

still simplifying

Answer 14

I got this
\[\frac{u^5}{10}+\frac{u^3}{3}-\frac{2014u^3}{3}+\frac{1}{2}u-2014u+C \text{ where } \tan(x/2)=u\]
so far
seems ugly

Answer 15

I wonder if I made a mistake because I don't have any really really large exponents

Answer 16

Something ain't right. There should be at least one large exponent in the neighborhood of 2014.

Answer 17

oops I see what i did
i mistaked (1-u^2) with (1+u^2)
looked at the signs too fast when simplifying

Answer 18

so
we have this with tan(x/2)=u
so far:
\[2 \int\limits_{}^{}\frac{(u^2+1)^3(1+u^2)^{2014}-4u^2 \cdot 2014 (u^2+1)^{2015}}{4u^2(1-u^2)^{2014}} du\]

Answer 19

which i don't see how we could make this pretty

Answer 20

The first two expressions in the numerator can be simplified to (u^2+1)^2017

Answer 21

well i know that but i'm saying after the easy stuff lol

Answer 22

Out of the trigonometric frying pan into the algebraic fire.

Answer 23

I will be right back.
I have to eat.
I will think about the problem.

Answer 24

Okay, thanks.

Answer 25

\[
\int \frac{cosec^2(x)-2014}{cos^{2014}(x)}dx = \int \frac{cosec^2(x)}{cos^{2014}(x)}dx - 2014\int \frac{1}{cos^{2014}(x)}dx = \\
\int \frac{1}{sin^2(x) * cos^{2014}(x)}dx - 2014\int \frac{1}{cos^{2014}(x)}dx = \\
u = tan(\frac x2); ~~~~dx = \frac{2du}{1+u^2}; ~~~~ cos(x) = \frac{1-u^2}{1+u^2}; ~~~~ sin(x) = \frac{2u}{1+u^2}\\
\int \frac{(1+u^2)^{2015}}{2u^2(1-u^2)^{2014}}du - 4028 \int \frac{(1+u^2)^{2013}}{(1-u^2)^{2014}}du
\]

Answer 26

I think it is B.
We can do small numbers and try to find a pattern.
For example:
I think
\[\int\limits_{}^{}\frac{\csc^2(x)-n}{\cos^n(x)} dx=-\sec^{n}(x)\cot(x)+C ,\text{ where } n \text{ is a natural number }\]
We can try to prove this by induction.

Answer 27

lets do where n is a natural number greater than 2

Answer 28

so starting at n=2
\[\int\limits_{}^{}\frac{\csc^2(x)-2}{\cos^2(x)} dx=\int\limits_{}^{}\frac{(\frac{u^2+1}{2u})^2-2}{(\frac{1-u^2}{1+u^2})^2} 2(u^2+1) du \\ \int\limits_{}^{}\frac{\frac{(u^2+1)^2}{4u^2}-2}{\frac{(1-u^2)^2}{(1+u^2)^2}} 2(u^2+1)du \\ \int\limits_{}^{} \frac{\frac{(u^2+1)^3}{2u^2}-4(u^2+1)}{\frac{(1-u^2)^2}{(1+u^2)^2} }du=\int\limits_{}^{}\frac{(u^2+1)^5-8u^2(u^2+1)^3}{2u^2 (1-u^2)^2} du\]

Answer 29

not finished

Answer 30

of course we could have done the other approach without this sub

Answer 31

\[\int\limits_{}^{}\frac{(u^2)^5+5(u^2)^4+10(u^2)^3+10(u^2)^2+5(u^2)+1-8u^2((u^2)^3+3(u^2)^2+3(u^2)+1)}{2u^2(1-2u^2+u^4)}du\]
\[\int\limits_{}^{}\frac{u^{10}+5u^8+10u^6+10u^4+5u^2+1-8u^8-24u^6-24u^4-8u^2}{2u^2(u^4-2u^2+1)} du\]
\[\int\limits_{}^{}\frac{u^{10}-3u^8-14u^6-14u^4-3u^2+1}{2u^6-4u^4+2u^2} du\]

Answer 32

division...

Answer 33

Way too much typing...

Answer 34

this problem is crazy. where did it come from?

Answer 35

I will do it the other way...

Answer 36

\[\int\limits_{}^{}\frac{\csc^2(x)-2}{\cos^2(x)}dx=\int\limits_{}^{}\frac{\csc^2(x)}{\cos^2(x)} dx -\int\limits_{}^{}\frac{2}{\cos^2(x)} dx \\ =\int\limits_{}^{}\frac{1}{\sin^2(x) \cos^2(x)} dx-\int\limits_{}^{}2 \sec^2(x) dx \\ =\int\limits_{}^{}\frac{\sin^2(x)+\cos^2(x)}{\sin^2(x)\cos^2(x)} dx-2 \tan(x)+C \]
\[=\int\limits_{}^{}\sec^2(x) dx+\int\limits_{}^{}\csc^2(x) dx-2 \tan(x)+C\]
\[=\tan(x)-\cot(x)-2\tan(x)+C=-\tan(x)-\cot(x)+C=\frac{-\sin(x)}{\cos(x)}+\frac{-\cos(x)}{\sin(x)}+C \\ = \frac{-\sin^2(x)-\cos^2(x)}{\cos(x)\sin(x)} +C =\frac{-1}{\cos(x)\sin(x)}+C=-\frac{\cos(x)}{\cos^2(x) \sin(x)}+C \\ =-\sec^2(x) \cot(x) +C\]

Answer 37

yeah that was much easier

Answer 38

for we just proved the first case

Answer 39

now if we assume \[\int\limits_{}^{}\frac{\csc^2(x)-k}{\cos^{k}(x)} dx=-\sec^{k}(x) \cot(x) +C \text{ for some integer } k \ge 2 \]
can we show \[\int\limits_{}^{}\frac{\csc^2(x)-k-1}{\cos^{k+1}(x) } dx = -\sec^{k+1} \cot(x) +C \]
let's play and try to show this.
\[\int\limits_{}^{}\frac{\csc^2(x)-k}{\cos^{k+1}(x)} dx -\int\limits_{}^{} \frac{1}{\cos^{k+1}(x)} dx\]
\[=\int\limits_{}^{}\frac{\csc^2(x)-k}{\cos^k(x)}\sec(x)dx-\int\limits_{}^{}\sec^{k+1} (x) dx \\ =-\sec^k(x)\cot(x)\sec(x)+\int\limits_{}^{}\sec^k(x)\cot(x)\sec(x)\tan(x) dx-\int\limits_{}^{}\sec^{k+1}(x) dx \\ =-\sec^{k+1}(x)\cot(x)+\int\limits_{}^{}\sec^{k+1} dx-\int\limits_{}^{}\sec^{k+1}(x) dx+C \\ =-\sec^{k+1}(x)\cot(x)+C \]

Answer 40

Therefore we have shown
\[\int\limits_{}^{}\frac{\csc^2(x)-n}{\cos^n(x)} dx =-\sec^n(x) \cot(x)+C \text{ for all natural } n \ge 2 \]

Answer 41

by way i did use integration by parts in my inductive step

Answer 42

Thank you @freckles. Very nicely done! This was in one of the tests for a college entrance examination that my nephew in India took last year.

Answer 43

It is a tough question for an entrance exam I would think.
But yeah I think the best thing to do here is start with small numbers and see what the pattern is.

Answer 44

Wolfram was choking not only on this integration but also on differentiating 3 out of the 4 answer choices!

There is another interesting non-calculus question that was also asked in the same entrance exam which i will post next.

Answer 45

I will totally take a look.

Answer 46

omg ranga I just realized something
we made that previous problem harder than it was I think...

Answer 47

\[\int\limits_{}^{} \frac{\csc^2(x)-2014}{\cos^{2014}(x)} dx \\ =\int\limits_{}^{}(\csc^2(x)-2014) \sec^{2014}(x) dx \\ \]
\[=\int\limits_{}^{}\csc^2(x) \sec^{2014}(x) dx- \int\limits_{}^{}2014 \sec^{2014}(x) dx \\ = \text{ ( integration by parts ) } \\ -\cot(x)\sec^{2014}(x)+\int\limits_{}^{}\cot(x)2014\sec^{2013}(x)\sec(x)\tan(x) dx-\int\limits_{}^{}2014 \sec^{2014}(x) dx\]
\[=-\cot(x)\sec^{2014}(x)+\int\limits_{}^{}2014\sec^{2014}(x)dx-\int\limits 2014\sec^{2014}(x) dx+C\]

Answer 48

I feel so dumb I didn't think of this earlier

Answer 49

giving us the answer we wanted
which was \[=-\cot(x)\sec^{2014}(x)+C \]

Answer 50

it was just a simple integration by parts question

Answer 51

OMG! That is so cool! Thanks again.

Answer 52

@ranga I can't believe wolfram couldn't do that.

Answer 53

that is an awesome question lol

Answer 54

This test is full of such awesome questions. Not too much labor intensive but real good, thinking type of questions. I can go on and on. The ones my nephew got stuck in I do it under non-test condition at my own leisurely pace and solve a majority of them. But some I get stuck and I don't want to spend too much time on just one problem.

BTW, I just posted the other algebra question from the same test.