A calculus problem by Jonathan Schirmer

Calculus Level 4

The following can be expressed as π + ln ⁡ a / b c \frac{\pi+\ln{a/b}}{c} where a a and b b are coprime positive integers:

∫ 0 π 4 d x 2 + tan ⁡ x \int^{\frac{\pi}{4}}_0 \frac{dx}{2+\tan{x}}

What is the value of a + b + c a+b+c ?


The answer is 27.

Jonathan Schirmer by Jonathan Schirmer , 27, USA

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2 solutions

Jing Hao Pang
Dec 16, 2013

∫ 0 π 4 d x 2 + tan ⁡ x \displaystyle\int_0^{\frac{\pi}{4}} \frac{dx}{2+\tan x}

= ∫ 0 π 4 sec ⁡ 2 x d x ( sec ⁡ 2 x ) ( 2 + tan ⁡ x ) = \displaystyle\int_0^{\frac{\pi}{4}} \frac{\sec^2 x \; dx}{(\sec^2 x)(2+\tan x)}

The reason to choose specifically sec ⁡ 2 x \sec^2 x is because of its close relation to tan ⁡ x \tan x :

Let u = tan ⁡ x u = \tan x

∙ d u d x = sec ⁡ 2 x ⇒ d u = sec ⁡ 2 x d x \bullet \frac{du}{dx} = \sec^2 x \Rightarrow du = \sec^2 x \; dx

∙ sec ⁡ 2 x = u 2 + 1 \bullet \sec^2 x= u^2 + 1

With this in mind, we can rewrite the expression to the following: ∫ 0 π 4 sec ⁡ 2 x d x ( sec ⁡ 2 x ) ( 2 + tan ⁡ x ) = ∫ 0 π 4 d u ( u 2 + 1 ) ( u + 2 ) \displaystyle\int_0^{\frac{\pi}{4}} \frac{\sec^2 x \; dx}{(\sec^2 x)(2+\tan x)} = \displaystyle\int_0^{\frac{\pi}{4}} \frac{du}{(u^2+1)(u+2)}

Applying partial fractions:

1 ( u 2 + 1 ) ( u + 2 ) = A u + B u 2 + 1 + C u + 2 \frac{1}{(u^2+1)(u+2)} = \frac {Au+B}{u^2+1} + \frac {C}{u+2}

1 = ( A u + B ) ( u + 2 ) + C ( u 2 + 1 ) 1 = (Au+B)(u+2) + C(u^2+1)

Subbing different values of u u , we get A = − 1 5 , B = 2 5 , C = 1 5 A=-\frac{1}{5}, B=\frac{2}{5}, C=\frac{1}{5}

Thus:

1 ( u 2 + 1 ) ( u + 2 ) = 2 − u 5 ( u 2 + 1 ) + 1 5 ( u + 2 ) \frac{1}{(u^2+1)(u+2)} = \frac {2-u}{5(u^2+1)} + \frac {1}{5(u+2)}

∫ 0 π 4 ( 2 − u 5 ( u 2 + 1 ) + 1 5 ( u + 2 ) ) d u \displaystyle\int_0^{\frac{\pi}{4}} (\frac {2-u}{5(u^2+1)} + \frac {1}{5(u+2)}) du

= 1 5 ∫ 0 π 4 2 − u ( u 2 + 1 ) d u + 1 5 ∫ 0 π 4 1 u + 2 d u = \displaystyle\frac{1}{5} \int_0^{\frac{\pi}{4}} \frac {2-u}{(u^2+1)} du + \frac{1}{5}\int_0^{\frac{\pi}{4}} \frac {1}{u+2} du

1 5 ∫ 0 π 4 2 − u ( u 2 + 1 ) d u \displaystyle\frac{1}{5} \int_0^{\frac{\pi}{4}} \frac {2-u}{(u^2+1)} du

= 1 5 ∫ 0 π 4 ( 2 − tan ⁡ x ) ( sec ⁡ 2 x ) d x ( sec ⁡ 2 x ) =\displaystyle\frac{1}{5} \int_0^{\frac{\pi}{4}} \frac {(2-\tan x)(\sec^2 x) \; dx}{(\sec^2 x)}

= 1 5 ∫ 0 π 4 ( 2 − tan ⁡ x ) d x =\displaystyle\frac{1}{5} \int_0^{\frac{\pi}{4}} (2-\tan x) \; dx

= 1 5 ( 2 x + ln ⁡ ∣ cos ⁡ x ∣ ) ∣ 0 π 4 =\displaystyle\frac{1}{5} (2x+\ln |\cos x|)\bigg|_0^{\frac{\pi}{4}}

= π 2 + ln ⁡ ( 1 2 ) 5 =\displaystyle\frac{\frac{\pi}{2} + \ln (\frac{1}{\sqrt{2}})}{5}

1 5 ∫ 0 π 4 1 u + 2 d u \displaystyle\frac{1}{5}\int_0^{\frac{\pi}{4}} \frac {1}{u+2} du

= l n ( u + 2 ) 5 ∣ 0 π 4 =\displaystyle\frac{ln (u+2)}{5}\Bigg|_0^{\frac{\pi}{4}}

= l n ( tan ⁡ x + 2 ) 5 ∣ 0 π 4 =\displaystyle\frac{ln (\tan x+2)}{5}\Bigg|_0^{\frac{\pi}{4}}

= l n 3 2 5 =\displaystyle\frac{ln \frac{3}{2}}{5}

Adding the two terms together:

π 2 + ln ⁡ ( 1 2 ) 5 + l n 3 2 5 \displaystyle\frac{\frac{\pi}{2} + \ln (\frac{1}{\sqrt{2}})}{5} + \frac{ln \frac{3}{2}}{5}

= π 2 + ln ⁡ ( 3 2 2 ) 5 =\displaystyle\frac{\frac{\pi}{2} + \ln (\frac{3}{2\sqrt{2}})}{5}

= π + 2 ln ⁡ ( 3 2 2 ) 10 =\displaystyle\frac{\pi + 2\ln (\frac{3}{2\sqrt{2}})}{10}

= π + ln ⁡ ( 9 8 ) 10 =\displaystyle\frac{\pi + \ln (\frac{9}{8})}{10}

Therefore, a + b + c = 9 + 8 + 10 = 27 a+b+c = 9+8+10 = \boxed {27}

11 Helpful 0 Interesting 1 Brilliant 0 Confused

Hi Jing! Thanks for a good solution! I would suggest in the Best of Calculus feed try reading Nicolae's post and then you'll see how magically 'less work' you could do and solve the integral!. https://brilliant.org/discussions/thread/an-interesting-trigonometric-integral/

Jit Ganguly - 7 years, 5 months ago

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Wow, I had a feeling that there's a shorter method ><. Thanks for the info! I'll shall input the alternative solution based on Nicolae's post here:

Let I = ∫ 0 π 4 1 2 + tan ⁡ x d x \displaystyle I = \int_0^\frac{\pi}{4} \frac {1}{2+\tan x} dx

= ∫ 0 π 4 1 2 + sin ⁡ x cos ⁡ x d x \displaystyle = \int_0^\frac{\pi}{4} \frac {1}{2+\frac{\sin x}{\cos x}} dx

= ∫ 0 π 4 cos ⁡ x sin ⁡ x + 2 cos ⁡ x d x \displaystyle = \int_0^\frac{\pi}{4} \frac {\cos x}{\sin x+2\cos x} dx

Let J = ∫ 0 π 4 sin ⁡ x sin ⁡ x + 2 cos ⁡ x d x \displaystyle J = \int_0^\frac{\pi}{4} \frac {\sin x}{\sin x+2\cos x} dx

Thus,

2 I + J = ∫ 0 π 4 sin ⁡ x + 2 cos ⁡ x sin ⁡ x + 2 cos ⁡ x d x 2I + J = \displaystyle \int_0^\frac{\pi}{4} \frac {\sin x+2\cos x}{\sin x+2\cos x} dx

= ∫ 0 π 4 sin ⁡ x + 2 cos ⁡ x sin ⁡ x + 2 cos ⁡ x d x = \displaystyle \int_0^\frac{\pi}{4} \frac {\sin x+2\cos x}{\sin x+2\cos x} dx

= x ∣ 0 π 4 = \displaystyle x|_0^\frac{\pi}{4}

= π 4 =\displaystyle \frac{\pi}{4}

I − 2 J = ∫ 0 π 4 cos ⁡ x − 2 sin ⁡ x sin ⁡ x + 2 cos ⁡ x d x I - 2J = \displaystyle \int_0^\frac{\pi}{4} \frac {\cos x-2\sin x}{\sin x+2\cos x} dx

Since d d x sin ⁡ x + 2 cos ⁡ x = cos ⁡ x − 2 sin ⁡ x \displaystyle \frac{d}{dx} \sin x + 2\cos x = \cos x - 2\sin x , the fraction is in the form of f ′ ( x ) f ( x ) \displaystyle \frac{f'(x)}{f(x)} , the integral becomes:

I − 2 J = ln ⁡ ∣ sin ⁡ x + 2 cos ⁡ x ∣ ∣ 0 π 4 I - 2J = \displaystyle \ln|\sin x + 2\cos x|\Big |_0^\frac{\pi}{4}

= ln ⁡ 3 2 − ln ⁡ 2 = \displaystyle \ln \frac{3}{\sqrt{2}} - \ln 2

= ln ⁡ 3 2 2 = \displaystyle \ln \frac{3}{2\sqrt{2}}

Now, simply solve the simultaneous equation to get

I = π + ln ⁡ ( 9 8 ) 10 \displaystyle I=\frac{\pi + \ln (\frac{9}{8})}{10}

Therefore, a + b + c = 9 + 8 + 10 = 27 \displaystyle a+b+c = 9+8+10 = \boxed{27}

Jing Hao Pang - 7 years, 5 months ago

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thats the same way i approached the problem!! cheers!!

G C KEERTHI Vasan - 7 years, 4 months ago

thats how i did it

Vinit Kumar - 7 years, 4 months ago

exactly as i did..

Shubham Jain - 7 years, 4 months ago

That's exactly how I did it, thanks to Nicolae. :)

Pranav Arora - 7 years, 5 months ago

why make the solutions so big and thoughtful ...... instead why cant you justtransfrom to sinx cosx form and then substitute sinx=2tanx/2 /!+tan^2x/2

arafat khan - 7 years, 5 months ago

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Could you further elaborate on your method? I'm open to different and more elegant solutions :)

Jing Hao Pang - 7 years, 5 months ago

We want: ∫ 0 π 4 c o s x 2 c o s x + s i n x \int_0^{\frac{\pi}{4}} \frac{cosx}{2cosx+sinx}

Let I 1 = ∫ 0 π 4 c o s x 2 c o s x + s i n x I_1 = \int_0^{\frac{\pi}{4}} \frac{cosx}{2cosx+sinx} and I 2 = ∫ 0 π 4 s i n x 2 c o s x + s i n x I_2 = \int_0^{\frac{\pi}{4}} \frac{sinx}{2cosx+sinx}

2 I 1 + I 2 = ∫ 0 π 4 d x = π 4 2I_1 + I_2 = \int_0^{\frac{\pi}{4}} dx = \frac{\pi}{4}

− 2 I 2 + I 1 = ∫ 0 π 4 − 2 s i n x + c o s x 2 c o s x + s i n x d x = l n ( 2 c o s x + s i n x ) ∣ 0 π 4 = l n 3 2 2 -2I_2 + I_1 = \int_0^{\frac{\pi}{4}} \frac{-2sinx+cosx}{2cosx+sinx} dx = ln(2cosx + sinx) |_0^{\frac{\pi}{4}} = ln \frac{3}{2\sqrt{2}}

5 I 1 = 2 ∗ π 4 + l n 3 2 2 = π + l n 9 8 2 5I_1 = 2 * \frac{\pi}{4} + ln \frac{3}{2\sqrt{2}} = \frac{\pi + ln \frac{9}{8}}{2}

∫ 0 π 4 c o s x 2 c o s x + s i n x = π + l n 9 8 10 \int_0^{\frac{\pi}{4}} \frac{cosx}{2cosx+sinx} = \frac{\pi + ln \frac{9}{8}}{10}

a + b + c = 9 + 8 + 10 = 27 a+b+c=9+8+10=27

7 Helpful 0 Interesting 0 Brilliant 0 Confused

Wonderfully done....

S Aditya - 4 years, 4 months ago

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