જો int left( frac{1-5 cos^{2}x}{sin^{5}x cos^ | vedclass

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Question

(B) ધારો કે $I = \int \left( \frac{1-5 \cos^{2}x}{\sin^{5}x \cos^{2}x} \right) dx = \int \left( \frac{\sec^{2}x}{\sin^{5}x} - \frac{5}{\sin^{5}x} \rig

Vedclass · Accepted Answer

$\frac{4}{\sqrt{3}}(8-\sqrt{6})$

Vedclass · Answer

(B) ધારો કે $I = \int \left( \frac{1-5 \cos^{2}x}{\sin^{5}x \cos^{2}x} ight) dx = \int \left( \frac{\sec^{2}x}{\sin^{5}x} - \frac{5}{\sin^{5}x} ight) dx$. $\int \frac{\sec^{2}x}{\sin^{5}x} dx$ પર ખંડશઃ સંકલન (Integration by Parts) વાપરતા, $u = \frac{1}{\sin^{5}x}$ અને $dv = \sec^{2}x dx$ લો. તેથી $du = -5 \sin^{-6}x \cos x dx$ અને $v = 	an x$. $\int \frac{\sec^{2}x}{\sin^{5}x} dx = \frac{	an x}{\sin^{5}x} - \int 	an x (-5 \sin^{-6}x \cos x) dx = \frac{	an x}{\sin^{5}x} + 5 \int \frac{\sin x / \cos x \cdot \cos x}{\sin^{6}x} dx = \frac{	an x}{\sin^{5}x} + 5 \int \frac{1}{\sin^{5}x} dx$. આ કિંમત $I$ માં મૂકતા: $I = \left( \frac{	an x}{\sin^{5}x} + 5 \int \frac{1}{\sin^{5}x} dx ight) - 5 \int \frac{1}{\sin^{5}x} dx = \frac{	an x}{\sin^{5}x} + C$. આમ, $f(x) = \frac{	an x}{\sin^{5}x} = \frac{1}{\cos x \sin^{4}x}$. $f(\frac{\pi}{6}) = \frac{1}{\cos(\pi/6) \sin^{4}(\pi/6)} = \frac{32}{\sqrt{3}}$. $f(\frac{\pi}{4}) = \frac{1}{\cos(\pi/4) \sin^{4}(\pi/4)} = 4\sqrt{2}$. $f(\frac{\pi}{6}) - f(\frac{\pi}{4}) = \frac{32}{\sqrt{3}} - 4\sqrt{2} = \frac{4}{\sqrt{3}}(8 - \sqrt{6})$.

જો $\int \left( \frac{1-5 \cos^{2}x}{\sin^{5}x \cos^{2}x} \right) dx = f(x) + C$ જ્યાં $C$ એ સંકલનનો અચળાંક છે, તો $f(\frac{\pi}{6}) - f(\frac{\pi}{4})$ ની કિંમત શોધો.

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