underset(-pi//4)overset(pi//4)int (dx)/(...

`underset(-pi//4)overset(pi//4)int (dx)/(1+cos 2x)` बराबर है

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The correct Answer is:

underset(0)overset(1)int (dx)/(e^(x)+e^(-x))
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underset(0)overset(1)int (dx)/(e^(x)+e^(-x))
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underset(1)overset(2)int (dx)/(sqrt((x-1)(2-x)))
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underset(0)overset(1)int (x dx)/(sqrt(1+x^(2)))
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underset(0)overset(pi)int x sin x cos^(2)x dx
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underset(0)overset(1//2)int (1)/((1+x^(2))sqrt(1-x^(2)))dx
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underset(0)overset(2)int (x^(2)+3)dx
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underset(0)overset(2)int e^(x)dx
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underset(0)overset(1)int x log(x+2)dx
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underset(0)overset(pi)int x log sin x dx
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underset(-pi//4)overset(pi//4)int log(sin x+cos x)dx
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underset(0)overset(pi)int (x)/(1+sin x)dx
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underset(pi//3)overset(pi//2)int (sqrt(1+cosx))/((1-cos x)^(5//2))dx
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underset(0)overset(pi//2)int (dx)/((a^(2) cos^(2)x+b^(2)sin^(2)x)^(2))
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underset(0)overset(pi//2)int sqrt(1+sin 2xdx) का मान है
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underset(-pi//4)overset(pi//4)int (dx)/(1+cos 2x) बराबर है
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underset(0)overset(pi//2)int cos x.e^(sin x)dx=...
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underset(-pi)overset(pi)int sin^(3)x cos^(2) x dx=......
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यदि underset(0)overset(a)int (1)/(1+4x^(2))dx=pi/8, "तो"a.......
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