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Calculus Level 4

∫ 0 π / 2 8 sin ⁡ x cos ⁡ x 3 + cos ⁡ 4 x d x = ? \large \int_0^{\pi/2} \dfrac{8\sin x \cos x}{3+\cos 4x} \, dx = \, ?

π 6 \dfrac{\pi}6 π 2 \dfrac{\pi}2 π 3 \dfrac{\pi}3 π 4 \dfrac{\pi}4

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Mohammad Hamdar by Mohammad Hamdar , 22, Lebanon

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2 solutions

∫ 0 π 2 8 sin ⁡ x cos ⁡ x 3 + cos ⁡ ( 4 x ) d x = ∫ 0 π 2 4 sin ⁡ ( 2 x ) 3 + 2 cos ⁡ 2 ( 2 x ) − 1 d x = ∫ 0 π 2 2 sin ⁡ ( 2 x ) 1 + cos ⁡ 2 ( 2 x ) d x Let tan ⁡ θ = cos ⁡ ( 2 x ) ⇒ sec ⁡ 2 θ d θ = − 2 sin ⁡ ( 2 x ) d x = ∫ π 4 − π 4 − sec ⁡ 2 θ 1 + tan ⁡ 2 θ d θ Note that 1 + tan ⁡ 2 θ = sec ⁡ 2 θ = ∫ − π 4 π 4 sec ⁡ 2 θ sec ⁡ 2 θ d θ = ∫ − π 4 π 4 d θ = [ θ ] − π 4 π 4 = π 2 \begin{aligned} \int_0^\frac{\pi}{2} \frac{8\sin x \cos x}{3 + \cos (4x)} dx & = \int_0^\frac{\pi}{2} \frac{4\sin (2x)}{3 + 2\cos^2 (2x) -1 } dx \\ & = \int_0^\frac{\pi}{2} \frac{2\sin (2x)}{1 + \cos^2 (2x)} dx \quad \quad \small \color{#3D99F6}{\text{Let} \tan \theta = \cos (2x) \quad \Rightarrow \sec^2 \theta \space d \theta = - 2\sin (2x) \space dx} \\ & = \int_{\frac{\pi}{4}}^{\color{#D61F06}{-\frac{\pi}{4}}} \frac{\color{#D61F06}{-} \sec^2 \theta}{1+\tan^2 \theta} d\theta \quad \quad \space \space \small \color{#3D99F6}{\text{Note that }1+\tan^2 \theta = \sec^2 \theta} \\ & = \int_{\color{#D61F06}{-\frac{\pi}{4}}}^{\frac{\pi}{4}} \frac{\sec^2 \theta}{\sec^2 \theta} d\theta = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} d\theta = \left[ \theta \right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = \boxed{\frac{\pi}{2}} \end{aligned}

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can you explain a little better how do you obtain sec^2(t) from 1+cos^2(2x) with the substitution tan(t)=cos(2x), ( i am referring to the denominator in the second line)please?

And can you also explain the reason why the minus sign in the substitution disappear? Should not be [-( sec^2(t)/sec^2(t))]?

guido barta - 5 years, 5 months ago

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I have added a line and red markings to explain your two questions. Hope that they help.

Chew-Seong Cheong - 5 years, 5 months ago

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thank you..now i understand

guido barta - 5 years, 5 months ago
Mohammad Hamdar
Jan 4, 2016

I = ∫ 0 π 2 4 sin ⁡ 2 x 3 + 2 cos ⁡ 2 2 x − 1 = − ∫ 0 π 2 − 2 sin ⁡ 2 x 1 + cos ⁡ 2 2 x k n o w i n g t h a t ( cos ⁡ 2 x ) ′ = − 2 sin ⁡ 2 x s o , I = { π 2 0 − arctan ⁡ ( cos ⁡ 2 x ) = π 2 I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { 4\sin { 2x } }{ 3+2\cos ^{ 2 }{ 2x\quad -1 } } } \quad \\ =-\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { -2\sin { 2x } }{ 1+\cos ^{ 2 }{ 2x } } } \quad \quad \quad \quad \quad \quad \quad knowing\quad that\quad (\cos { 2x)'=-2\sin { 2x } \quad } \\ so,\quad I=\begin{cases} \frac { \pi }{ 2 } \\ 0 \end{cases}-\arctan { (\cos { 2x } } )\quad =\quad \frac { \pi }{ 2 }

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