Dear student, Let I = ∫xa2-x2a2+x2dxPut x2 = a2 cosy⇒2x dx = -a2sin y dyNow, I = -a22∫a2 - a2 cos ya2 + a2 cos y × sin y dy=-a22∫1 - cos y1 + cos y 2 tany/21+tan2y/2 dy=-a2∫ tan2y/2sec2y/2 dy=-a2∫sin2y/2 dy=-a2∫1 - cos y2 dy=-a22y - sin y=-a22y - 1 - cos2y + C=-a22cos-1x2/a2 - 1 - x4a4 + C=-a22π2 - sin-1x2a2 - a4 - x4a2 + C=-πa24 + a22sin-1x2a2 + 12a4-x4 + C=a22sin-1x2a2 + 12a4-x4 + C Hope this information will clear your doubts If you have any more doubts just ask here on the forum and our experts will try to help out as soon as possible Regards

pls solve Qno.9
pls solve Qno.9 x

Dear student,

Let I = \int x \sqrt{\frac{a^{2} - x^{2}}{a^{2} + x^{2}}} dx Put x^{2} = a^{2} cosy \Rightarrow 2 x dx = - a^{2} \sin y dy Now, I = \frac{- a^{2}}{2} \int \sqrt{\frac{a^{2} - a^{2} \cos y}{a^{2} + a^{2} \cos y}} \times \sin y dy = \frac{- a^{2}}{2} \int \sqrt{\frac{1 - \cos y}{1 + \cos y}} . \frac{2 \tan (y / 2)}{1 + \tan^{2} (y / 2)} dy = - a^{2} \int \frac{\tan^{2} (y / 2)}{\sec^{2} (y / 2)} dy = - a^{2} \int \sin^{2} (y / 2) dy = - a^{2} \int (\frac{1 - \cos y}{2}) dy = \frac{- a^{2}}{2} [y - \sin y] = - \frac{a^{2}}{2} [y - \sqrt{1 - \cos^{2} y}] + C = - \frac{a^{2}}{2} [\cos^{- 1} (x^{2} / a^{2}) - \sqrt{1 - \frac{x^{4}}{a^{4}}}] + C = - \frac{a^{2}}{2} [\frac{π}{2} - \sin^{- 1} (\frac{x^{2}}{a^{2}}) - \frac{\sqrt{a^{4} - x^{4}}}{a^{2}}] + C = - \frac{{πa}^{2}}{4} + \frac{a^{2}}{2} \sin^{- 1} (\frac{x^{2}}{a^{2}}) + \frac{1}{2} \sqrt{a^{4} - x^{4}} + C = \frac{a^{2}}{2} \sin^{- 1} (\frac{x^{2}}{a^{2}}) + \frac{1}{2} \sqrt{a^{4} - x^{4}} + C

Hope this information will clear your doubts. If you have any more doubts just ask here on the forum and our experts will try to help out as soon as possible.
Regards

View Full Answer

pls solve Qno.9​​pls solve Qno.9​​ x

pls solve Qno.9
pls solve Qno.9 x