Geometry or calculus?

Calculus Level 3

∫ 0 ∘ 53 ∘ ( tan ⁡ x 1 − cot ⁡ x + cot ⁡ x 1 − tan ⁡ x − 1 ) sin ⁡ 2 x cos ⁡ x d x \large \int_{{0}^{\circ}}^{{53}^{\circ}} \left(\dfrac{\tan x}{1 - \cot x} + \dfrac{\cot x}{1 - \tan x} - 1\right)\dfrac{{\sin}^2 x}{\cos x} \, dx

Evaluate the definite integral above to two decimal places.


The answer is 0.66.

Ashish Menon by Ashish Menon , 20, India

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1 solution

Ashish Menon
Jun 21, 2016

( tan ⁡ x 1 − cot ⁡ x + cot ⁡ x 1 − tan ⁡ x − 1 ) sin ⁡ 2 x cos ⁡ x = ( sin ⁡ x cos ⁡ x 1 − cos ⁡ x sin ⁡ x + cos ⁡ x sin ⁡ x 1 − sin ⁡ x cos ⁡ x − 1 ) sin ⁡ x sin ⁡ x cos ⁡ x = ( sin ⁡ 2 x cos ⁡ x ( sin ⁡ x − cos ⁡ x ) + cos ⁡ 2 x sin ⁡ x ( cos ⁡ x − sin ⁡ x ) − 1 ) sin ⁡ x tan ⁡ x = ( sin ⁡ 3 x − cos ⁡ 3 x sin ⁡ x cos ⁡ x ( sin ⁡ x − cos ⁡ x ) − 1 ) sin ⁡ x tan ⁡ x = ( sin ⁡ 2 x + cos ⁡ 2 x + sin ⁡ x cos ⁡ x sin ⁡ x cos ⁡ x − 1 ) sin ⁡ x tan ⁡ x = ( 1 sin ⁡ x cos ⁡ x + 1 − 1 ) sin ⁡ x tan ⁡ x = csc ⁡ x sec ⁡ x sin ⁡ x tan ⁡ x = sec ⁡ x tan ⁡ x So, the question reduces to : − ∫ 0 ° 53 ° sec ⁡ x tan ⁡ x d x = sec ⁡ x ∣ 0 ° 53 ° = sec ⁡ ( 53 ° ) − sec ⁡ ( 0 ° ) = 0.66 \begin{aligned} \left(\dfrac{\tan x}{1 - \cot x} + \dfrac{\cot x}{1 - \tan x} - 1\right)\dfrac{{\sin}^2 x}{\cos x} & = \left(\dfrac{\frac{\sin x}{\cos x}}{1 - \frac{\cos x}{\sin x}} + \dfrac{\frac{\cos x}{\sin x}}{1 - \frac{\sin x}{\cos x}} - 1\right)\sin x \dfrac{\sin x}{\cos x}\\ \\ & = \left(\dfrac{{\sin}^2 x}{\cos x\left(\sin x - \cos x\right)} + \dfrac{{\cos}^2 x}{\sin x\left(\cos x - \sin x\right)} - 1\right)\sin x \tan x\\ \\ & = \left(\dfrac{{\sin}^3 x - {\cos}^3 x}{\sin x\cos x\left(\sin x - \cos x\right)} - 1\right)\sin x\tan x\\ \\ & = \left(\dfrac{{\sin}^2 x + {\cos}^2 x + \sin x\cos x}{\sin x\cos x} - 1\right)\sin x \tan x\\ \\ & = \left(\dfrac{1}{\sin x\cos x} + 1 - 1\right)\sin x\tan x\\ \\ & = \csc x\sec x\sin x\tan x\\ \\ & = \sec x\tan x\\ \\ \text{So, the question reduces to}:-\\ \\ \int_{{0}^°}^{{53}^°} \sec x\tan x \, dx & = \sec x{\Large \vert}_{{0}^°}^{{53}^°}\\ \\ & = \sec({53}^°) - \sec({0}^°)\\ \\ & = \color{#3D99F6}{\boxed{0.66}} \end{aligned}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

One more thing is about the 'upper limit' of the integral , that you should either mention 5 3 ∘ 53^{\circ}

O R \large \boxed{OR}

You can write it in radians as 53 π 180 \dfrac{53 \pi}{180} , please do correct it !

Rishabh Tiwari - 4 years, 11 months ago

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Well, thanks for the correction, but keep in mind that if nothing is mentioned you should consider radians ;)

Ashish Menon - 4 years, 11 months ago

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Right ! I agree ; but sec ⁡ 5 3 ∘ \sec 53^{\circ} has the value of 5 3 \dfrac {5}{3} and not sec ⁡ 53 \sec 53 , right ? ; I.e.

sec ⁡ 5 3 ∘ = 5 3 = sec ⁡ 53 π 180 ≠ sec ⁡ 53 \sec 53^{\circ} = \dfrac {5}{3} = \sec {\dfrac {53 \pi}{180}} \neq \sec {53} , so in the last step , the evaluation will be wrong, Right? You might want to recorrect your solution a bit ? ;-)

Rishabh Tiwari - 4 years, 11 months ago

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@Rishabh Tiwari – Oh well the degree sign kk, that was understood but i will modify that typo :) And sec ⁡ 53 ° \sec{53}^° is not exactly equal to 3 5 \dfrac{3}{5} its an approximation :P

Ashish Menon - 4 years, 11 months ago

Nice solution! (+1) , btw isn't there a typo in the 4th line ?

It should be sin ⁡ 2 x + cos ⁡ 2 x + sin ⁡ x cos ⁡ x \sin^2 x + \cos^2 x + \sin x \cos x :-)

Rishabh Tiwari - 4 years, 11 months ago

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Well I corretced that. Thanks!

Ashish Menon - 4 years, 11 months ago

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