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∫dx/sin2x+tan2x ...
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\hspace{-16}\bf{(1)\;\; \int\sqrt{a+\sqrt{b+\sqrt{x}}}\; dx}$\\\\\\ $\bf{(2)\;\;\int \sqrt{1+2\sqrt{x-x^2}}\;dx}$ ...
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\hspace{-16}\bf{(1)\;\; \int\frac{1}{(1+x^4)^{\frac{1}{4}}}dx}$\\\\\\ $\bf{(2)\;\; \int\frac{1}{(1-x^4)^{\frac{1}{4}}}dx}$\\\\\\ $\bf{(3)\;\;\int\frac{1}{(1+x^4)}dx}$\\\\\\ $\bf{(4)\;\;\int\frac{1}{(1+x^6)}dx}$ ...
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f(x) = ∫ 1-sin(2x) then f(Π/4) = ? will it be 2 or 0. because in the sheet that you gave the answer given at the back is (cosx + sinx)sgn(cosx-sinx)+c ...
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± ...
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int {frac{{a{x^3} + b{x^2} + c}}{{{x^4}}}} dx equals to ...
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int_0^{1.5} {[{x^2}]dx} , where [.] denotes the greatest integer function, equals ...
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Let a, b, c bc non-zero real numbers such that int_{,0}^{,3} {} (3,a{x^2} + 2,bx + c),,dx = int_{,1}^{,3} {} (3,a{x^2} + 2,bx + c),,dx, then ...
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int {frac{{d heta }}{{sin heta ,.,{{cos }^3} heta }}}Â = ...
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int {{a^x}da = } ...
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int{cos 2x.sin 4xdx} equals ...
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int {frac{{dx}}{{sin x + sin 2x}}}Â = ...
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int {frac{{dx}}{{{{(x - 1)}^2}(x - 2)}}}Â = ...
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int {frac{{dx}}{{4{{sin }^2}x + 5{{cos }^2}x}}} ...