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Question Number 137003 by Mathspace last updated on 28/Mar/21
find ∫_0 ^(π/4) ((cos^5 t)/(cos(5t)))dt
$${find}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{cos}^{\mathrm{5}} {t}}{{cos}\left(\mathrm{5}{t}\right)}{dt} \\ $$
Answered by bobhans last updated on 29/Mar/21
cos (5t)=cos^5 t−10sin^2 t cos^3 t+5sin^4 t cos t cos (5t)=cos^5 t−10cos^3 t(1−cos^2 t)+5cos t(1−cos^2 t)^2 = cos^5 t−10cos^3 t+10cos^5 t+5cos t(1−2cos^2 t+cos^4 t) =11cos^5 t−10cos^3 t+5cos t−10cos^3 t+5cos^5 t =16cos^5 t−20cos^3 t+5cos t I=∫ ((cos^4 t)/(16cos^4 −20cos^2 t+5)) dt
$$\mathrm{cos}\:\left(\mathrm{5t}\right)=\mathrm{cos}\:^{\mathrm{5}} \mathrm{t}−\mathrm{10sin}\:^{\mathrm{2}} \mathrm{t}\:\mathrm{cos}\:^{\mathrm{3}} \mathrm{t}+\mathrm{5sin}\:^{\mathrm{4}} \mathrm{t}\:\mathrm{cos}\:\mathrm{t} \\ $$$$\mathrm{cos}\:\left(\mathrm{5t}\right)=\mathrm{cos}\:^{\mathrm{5}} \mathrm{t}−\mathrm{10cos}\:^{\mathrm{3}} \mathrm{t}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{t}\right)+\mathrm{5cos}\:\mathrm{t}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{t}\right)^{\mathrm{2}} \\ $$$$=\:\mathrm{cos}\:^{\mathrm{5}} \mathrm{t}−\mathrm{10cos}\:^{\mathrm{3}} \mathrm{t}+\mathrm{10cos}\:^{\mathrm{5}} \mathrm{t}+\mathrm{5cos}\:\mathrm{t}\left(\mathrm{1}−\mathrm{2cos}\:^{\mathrm{2}} \mathrm{t}+\mathrm{cos}\:^{\mathrm{4}} \mathrm{t}\right) \\ $$$$=\mathrm{11cos}\:^{\mathrm{5}} \mathrm{t}−\mathrm{10cos}\:^{\mathrm{3}} \mathrm{t}+\mathrm{5cos}\:\mathrm{t}−\mathrm{10cos}\:^{\mathrm{3}} \mathrm{t}+\mathrm{5cos}\:^{\mathrm{5}} \mathrm{t} \\ $$$$=\mathrm{16cos}\:^{\mathrm{5}} \mathrm{t}−\mathrm{20cos}\:^{\mathrm{3}} \mathrm{t}+\mathrm{5cos}\:\mathrm{t} \\ $$$$\mathrm{I}=\int\:\frac{\mathrm{cos}\:^{\mathrm{4}} \mathrm{t}}{\mathrm{16cos}\:^{\mathrm{4}} −\mathrm{20cos}\:^{\mathrm{2}} \mathrm{t}+\mathrm{5}}\:\mathrm{dt} \\ $$$$ \\ $$

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