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junior college 1 | H2 Maths
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Date Posted: 2 years ago
Views: 396
∫ tan⁴ 2x dx
= ∫ (tan² 2x)² dx
= ∫ (sec² 2x - 1)(tan² 2x) dx
(Recall that 1 + tan² θ = sec² θ)
= ∫ (sec² 2x tan² 2x - tan² 2x) dx
= ∫ (½tan² 2x (2sec² 2x) - (sec² 2x - 1) ) dx
= ∫ (½(tan 2x)² (2sec² 2x) - ½(2sec² 2x) + 1) dx
Realise that differentiating tan 2x gives 2 sec² 2x so this is directly integrable.
so the first part of the expression is in the form
y² dy/dx , where y = tan 2x and dy/dx = 2sec² 2x
So this integrates back into ⅓y³
= ½(tan 2x)³ / 3 - ½tan 2x + x
= 1/6 tan³ 2x - ½tan 2x + x + c
= ∫ (tan² 2x)² dx
= ∫ (sec² 2x - 1)(tan² 2x) dx
(Recall that 1 + tan² θ = sec² θ)
= ∫ (sec² 2x tan² 2x - tan² 2x) dx
= ∫ (½tan² 2x (2sec² 2x) - (sec² 2x - 1) ) dx
= ∫ (½(tan 2x)² (2sec² 2x) - ½(2sec² 2x) + 1) dx
Realise that differentiating tan 2x gives 2 sec² 2x so this is directly integrable.
so the first part of the expression is in the form
y² dy/dx , where y = tan 2x and dy/dx = 2sec² 2x
So this integrates back into ⅓y³
= ½(tan 2x)³ / 3 - ½tan 2x + x
= 1/6 tan³ 2x - ½tan 2x + x + c
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How did u get to step 2?
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