Maths · Integration by substitution, by parts and by partial fractions
d x= \)
\( \int\left(\tan ^{10} x+\tan ^{12} x\right) d x= \)
- A. \( \frac{\tan ^{9} x}{9}+c \)
- B. \( \frac{\tan ^{13} x}{13}+c \)
- C. \( \frac{\tan ^{11} x}{11}+c \)
- D. \( \frac{\tan ^{8} x}{8}+c \)
Step-by-step solution
Factor out \(\tan^{10} x\) to get \(\tan^{10} x (1+\tan^2 x)\). Using \(1+\tan^2 x = \sec^2 x = \frac{d}{dx}(\tan x)\), substitute \(u = \tan x\), \(du = \sec^2 x dx\). The integral becomes \(\int u^{10} du = \frac{u^{11}}{11} + C = \frac{\tan^{11} x}{11} + C\).
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