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JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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Application +4 | -1
The value of ∫dxx+a2−x2\int \frac{d x}{x+\sqrt{a^2-x^2}}, is equal to
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Correct option: A
12sin⁡−1(xa)+12log⁡∣x+a2−x2∣+C1\frac{1}{2} \sin ^{-1}\left(\frac{x}{a}\right)+\frac{1}{2} \log \left|x+\sqrt{a^2-x^2}\right|+C_1
Option A (12sin⁡−1(xa)+12log⁡∣x+a2−x2∣+C1\frac{1}{2} \sin ^{-1}(\frac{x}{a})+\frac{1}{2} \log |x+\sqrt{a^2-x^2}|+C_1) is correct because it is the result of the integral after applying the trigonometric substitution x=asin⁡θx = a \sin \theta and simplifying.

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∫sin⁡5xcos⁡4xdx=\int \sin ^5 x \cos ^4 x d x= If f(x)f(x) is differentibale and ∫0t2xf(x)dx=25t5\int\limits_{0}^{t^2} x f(x) d x=\frac{2}{5} t^5 then f(425)f\left(\frac{4}{25}\right) equals If ∫1−7cos⁡2xsin⁡7xcos⁡2xdx=f(x)sin⁡7x+C\int \frac{1-7 \cos ^2 x}{\sin ^7 x \cos ^2 x} d x=\frac{f(x)}{\sin ^7 x}+C is equal to If ∫ex2(2−1x2)de=ex2f(x)+C\int {e}^{{x}^2}\left(2-\frac{1}{{x}^2}\right) {de}={e}^{{x}^2} {f}({x})+{C} and f(12)=2{f}\left(\frac{1}{2}\right)=2, then f(1)f(1) is equal to (where C{C} is an arbitrary constant) ∫x7(1−x2)5dx\int \frac{x^7}{\left(1-x^2\right)^5} d x is equal to: ∫ ax3+  bx2+cx4   dx\int \,\frac{{a{x^3} + \,\,b{x^2} + c}}{{{x^4}}}\;\,dx equals
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