A kind of N.C.E.R.T problem

Calculus Level 2

$\displaystyle \int_{0}^{\frac{\pi}{2}} sec( x - \frac{\pi}{6})sec( x - \frac{\pi}{3})dx$

I know it is not a good problem , but please try it once and post your different approaches

The answer is 2.197.

1 solution

U Z
Dec 9, 2014

$\displaystyle \int_{0}^{\frac{\pi}{2}} sec( x - \frac{\pi}{6})sec( x - \frac{\pi}{3})dx$

$= \displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{1}{cos(x - \frac{\pi}{6})cos(x - \frac{\pi}{3})}dx$

$= \displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{2\times\frac{1}{2}}{cos(x - \frac{\pi}{6})cos(x - \frac{\pi}{3})}dx$

$=\boxed{ 2\displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{sin[(x - \frac{\pi}{6}) - (x - \frac{\pi}{3})]}{cos(x - \frac{\pi}{6})cos(x - \frac{\pi}{3})}dx}$

$= 2\displaystyle \int_{0}^{\frac{\pi}{2}} tan(x - \frac{\pi}{6}) - tan(x - \frac{\pi}{3})dx$

$=\boxed{2ln3}$

@megh choksi Sir, this is indeed a good problem.

Anuj Shikarkhane - 6 years, 6 months ago

Please , please i am not a sir

Sir are @Sandeep Bhardwaj @Deepanshu Gupta @Aditya Raut @Ayush Verma and many more

U Z - 6 years, 6 months ago

Omg ! what did You say ... Please I'am not a Sir , But I'am an Foolish Guy which is very Small brain but enjoys The Beauty of Maths and Physics ! So Please Don't give me such Honour I'am not capable for it ! You r too good in front of me !

Deepanshu Gupta - 6 years, 6 months ago

@Deepanshu Gupta – No ,no i am not good at physics just do maths, indeed you are the too good in front of me

U Z - 6 years, 6 months ago

Same as I did! :)

Pranjal Jain - 6 years, 6 months ago

Nice , can you try a different approach of(the first problem we study in indefinite integration)

$\displaystyle \int \dfrac{sinx}{sinx + cosx}$

U Z - 6 years, 6 months ago

I guess this might work.

$sinx=\dfrac{2tan\frac{x}{2}}{1+tan^{2}\frac{x}{2}}$ and $cosx=\dfrac{1-tan^{2}\frac{x}{2}}{1+tan^{2}\frac{x}{2}}$

Substitute $tan\frac{x}{2}=p$

I am a little busy. You may carry on with this.

Pranjal Jain - 6 years, 6 months ago

@Pranjal Jain – Nice!

$\displaystyle \dfrac{sinx}{\sqrt{2}sin(\dfrac{\pi}{4} + x)}$

$\displaystyle \dfrac{sin( (\dfrac{\pi}{4} + x) - \dfrac{\pi}{4})}{\sqrt{2}sin(\dfrac{\pi}{4} + x)}$

$\displaystyle \int \dfrac{1}{2} - \dfrac{1}{2}cot(x + \dfrac{\pi}{4})$

U Z - 6 years, 6 months ago

@U Z – Nice Megh ! $=\int { \cfrac { \cfrac { 1 }{ 2 } (2\sin { x) } }{ \sin { x } +\cos { x } } } \\ \quad \\ =\quad \cfrac { 1 }{ 2 } \int { \cfrac { (\sin { x } +\cos { x } )-(\cos { x } -\sin { x } ) }{ \sin { x } +\cos { x } } } \\ \\ =\quad \cfrac { 1 }{ 2 } \int { 1 } \quad -\quad \quad \cfrac { 1 }{ 2 } \int { \cfrac { \cos { x } -\sin { x } }{ \sin { x } +\cos { x } } } \quad \quad \quad (\quad \because \quad \cfrac { f^{ ' }\left( x \right) }{ f\left( x \right) } )$ .

Deepanshu Gupta - 6 years, 6 months ago

@U Z – Oh! How can I forget this! You did this so easily!!

Pranjal Jain - 6 years, 6 months ago

A kind of N.C.E.R.T problem

The answer is 2.197.

1 solution

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