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Integrate : $\int x^5 \sec^2 (7x^6+9) dx$

$(a)\;\frac{1}{42} \tan (7x^6 +9)+c \qquad(b)\;\frac{1}{42} \cot (7x^6 +9)+c \qquad(c)\;\frac{1}{42} \sin (7x^6 +9)+c \qquad (d)\;None$

1 Answer

  • $\int sec^2x\:dx=tanx$
Given: $I=\int x^5 \sec^2 (7x^6+9) dx$
Let $7x^6+9=t$
Differentiating both the sides
$42 x^5 dx=dt$
Substituting the values of $t$ and $dx$ in I we get
$\Rightarrow\:I=\int \large\frac{1}{42} $$\sec^2t +c $
$\Rightarrow\:I=\large\frac{1}{42} $$\tan t+c $ (Since $\int sec^2t dt=tant$)
Substituting the value of $t$ we get
$\Rightarrow\:I=\large\frac{1}{42}$$ \tan (7x^6 +9)+c $
Hence (a) is the correct answer.
answered Dec 23, 2013 by meena.p
edited Mar 13, 2014 by rvidyagovindarajan_1