How to calculate $int_0^pi/4sin^n+1thetacos^n-1theta dtheta$?

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My approach is to use the properties of the beta function. But the limits of integration are from zero to $fracpi4$. We know that




$$
B(m,n) = 2int_0^pi/2sin^2m-1theta cos^2n-1theta dtheta = dfracGamma(m)Gamma(n)Gamma(m+n)
$$




Thanks for any help.




calculus integration definite-integrals special-functions




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  • $$sin^n+1tcos^n-1t=dfractan^n+1tsec^2tsec^2n+2t$$
    – lab bhattacharjee
    Jul 25 at 14:32















up vote
3
down vote

favorite
1












My approach is to use the properties of the beta function. But the limits of integration are from zero to $fracpi4$. We know that




$$
B(m,n) = 2int_0^pi/2sin^2m-1theta cos^2n-1theta dtheta = dfracGamma(m)Gamma(n)Gamma(m+n)
$$




Thanks for any help.




calculus integration definite-integrals special-functions




share|cite|improve this question





















  • $$sin^n+1tcos^n-1t=dfractan^n+1tsec^2tsec^2n+2t$$
    – lab bhattacharjee
    Jul 25 at 14:32













up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1





My approach is to use the properties of the beta function. But the limits of integration are from zero to $fracpi4$. We know that




$$
B(m,n) = 2int_0^pi/2sin^2m-1theta cos^2n-1theta dtheta = dfracGamma(m)Gamma(n)Gamma(m+n)
$$




Thanks for any help.




calculus integration definite-integrals special-functions




share|cite|improve this question













My approach is to use the properties of the beta function. But the limits of integration are from zero to $fracpi4$. We know that




$$
B(m,n) = 2int_0^pi/2sin^2m-1theta cos^2n-1theta dtheta = dfracGamma(m)Gamma(n)Gamma(m+n)
$$




Thanks for any help.





calculus integration definite-integrals special-functions





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edited Jul 30 at 12:52









Harry Peter

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asked Jul 25 at 14:24









MathFacts

2,9891424




2,9891424











  • $$sin^n+1tcos^n-1t=dfractan^n+1tsec^2tsec^2n+2t$$
    – lab bhattacharjee
    Jul 25 at 14:32

















  • $$sin^n+1tcos^n-1t=dfractan^n+1tsec^2tsec^2n+2t$$
    – lab bhattacharjee
    Jul 25 at 14:32
















$$sin^n+1tcos^n-1t=dfractan^n+1tsec^2tsec^2n+2t$$
– lab bhattacharjee
Jul 25 at 14:32





$$sin^n+1tcos^n-1t=dfractan^n+1tsec^2tsec^2n+2t$$
– lab bhattacharjee
Jul 25 at 14:32











1 Answer
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Hint: Use substitution $theta=dfracu2$ then continue with
$$dfrac12^n+1int_0^pi/2sin^n-1u(1-cos u)du=dfracbeta(fracn2,frac12)-beta(fracn2,1)2^n+2$$






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  • The so-called Weierstrass substitution works miracles with trigonometric integrals.
    – Pedro Tamaroff♦
    Jul 26 at 2:02










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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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oldest

votes








up vote
6
down vote













Hint: Use substitution $theta=dfracu2$ then continue with
$$dfrac12^n+1int_0^pi/2sin^n-1u(1-cos u)du=dfracbeta(fracn2,frac12)-beta(fracn2,1)2^n+2$$






share|cite|improve this answer





















  • The so-called Weierstrass substitution works miracles with trigonometric integrals.
    – Pedro Tamaroff♦
    Jul 26 at 2:02














up vote
6
down vote













Hint: Use substitution $theta=dfracu2$ then continue with
$$dfrac12^n+1int_0^pi/2sin^n-1u(1-cos u)du=dfracbeta(fracn2,frac12)-beta(fracn2,1)2^n+2$$






share|cite|improve this answer





















  • The so-called Weierstrass substitution works miracles with trigonometric integrals.
    – Pedro Tamaroff♦
    Jul 26 at 2:02












up vote
6
down vote










up vote
6
down vote









Hint: Use substitution $theta=dfracu2$ then continue with
$$dfrac12^n+1int_0^pi/2sin^n-1u(1-cos u)du=dfracbeta(fracn2,frac12)-beta(fracn2,1)2^n+2$$






share|cite|improve this answer













Hint: Use substitution $theta=dfracu2$ then continue with
$$dfrac12^n+1int_0^pi/2sin^n-1u(1-cos u)du=dfracbeta(fracn2,frac12)-beta(fracn2,1)2^n+2$$







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answered Jul 25 at 14:32









Nosrati

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  • The so-called Weierstrass substitution works miracles with trigonometric integrals.
    – Pedro Tamaroff♦
    Jul 26 at 2:02
















  • The so-called Weierstrass substitution works miracles with trigonometric integrals.
    – Pedro Tamaroff♦
    Jul 26 at 2:02















The so-called Weierstrass substitution works miracles with trigonometric integrals.
– Pedro Tamaroff♦
Jul 26 at 2:02




The so-called Weierstrass substitution works miracles with trigonometric integrals.
– Pedro Tamaroff♦
Jul 26 at 2:02












 

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