A) \[\frac{1}{3(1+{{\tan }^{3}}x)}+C\]
B) \[-\frac{1}{3(1+{{\tan }^{3}}x)}+C\]
C) \[\frac{1}{1+{{\tan }^{3}}x}+C\]
D) \[-\frac{1}{1+{{\tan }^{3}}x}+C\]
Correct Answer: B
Solution :
Let \[I=\int_{{}}^{{}}{\frac{{{\sin }^{2}}x{{\cos }^{2}}x}{{{({{\sin }^{3}}x+{{\cos }^{3}}x)}^{2}}}}dx\] \[\int_{{}}^{{}}{\frac{{{\tan }^{2}}x{{\sec }^{2}}x}{{{(1+{{\tan }^{3}}x)}^{2}}}dx}\] [dividing numerator and denominator by \[{{\cos }^{6}}x\]] \[=\frac{1}{3}\int_{{}}^{{}}{\frac{3{{\tan }^{2}}x{{\sec }^{2}}x}{{{(1+{{\tan }^{3}}x)}^{2}}}}dx\] \[=\frac{1}{3}\int_{{}}^{{}}{\frac{1}{{{(1+{{\tan }^{3}}x)}^{2}}}}d(1+{{\tan }^{3}}x)\] \[=-\frac{1}{3(1+{{\tan }^{2}}x)}+C\]You need to login to perform this action.
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