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The value of the integral $\int \frac{dx}{cos^{6}x+sin^{6}x}$ is equal to

Option 1)
log_e (tanx - cotx) + c

Option 2)
log_e (cotx - tanx) + c

Option 3)
tan^-1 (tanx - cotx) + c

Option 4)
none of these

Answers (1)

best_answer

As we learnt

Type of Integration by perfect square -

Integration in the form of

(i) $\int f(x+\frac{1}{x})(1-\frac{1}{x^{2}})dx$

(ii) $\int f(x-\frac{1}{x})(1+\frac{1}{x^{2}})dx$

(iii) $\int f(x^{2}+\frac{1}{x^{2}})(x-\frac{1}{x^{3}})dx$

(iv) $\int f(x^{2}-\frac{1}{x^{2}})(x+\frac{1}{x^{3}})dx$

(v) $\int \frac{(1\pm \frac{1}{x^{2}})dx}{x^{2}+\frac{1}{x^{2}}}$

(vi) $\int \frac{f(x)dx}{ax^{4}+2bx^{3}+cx^{2}+2bx+a}$

- wherein

(i) $\rightarrow$ put $(x+\frac{1}{x})=t$

(ii) $\rightarrow$ put $(x-\frac{1}{x})=t$

(iii) $\rightarrow$ put $(x^{2}+\frac{1}{x^{2}})=t$

(iv) $\rightarrow$ put $(x^{2}-\frac{1}{x^{2}})=t$

(v) $\rightarrow$ for $1+\frac{1}{x^{2}}$ put $x-\frac{1}{x}=t$

$\rightarrow$ for $1-\frac{1}{x^{2}}$ put $x+\frac{1}{x}=t$

(vi) $\rightarrow$ put $(x+\frac{1}{x})=t$ if $b\neq 0$

put $(x^{2}+\frac{1}{x^{2}})=t$ if $b= 0$

$Let\: \: I = \int {\frac{{dx}}{{{{\cos }^6}x + {{\sin }^6}x}}} =\int {\frac{{{{\sec }^6}x}}{{1 + {{\tan }^6}x}}\;dx}$

$= \int {\frac{{{{(1 + {{\tan }^2}x)}^2}{{\sec }^2}x\;dx}}{{1 + {{\tan }^6}x}}}$

If tanx = p, then sec²x dx = dp

$I = \int {\frac{{{{(1 + {p^2})}^2}dp}}{{1 + {p^6}}}} = \int {\frac{{(1 + {p^2})}}{{{p^4} - {p^2} + 1}}} \;dp\; = \int {\frac{{{p^2}\left( {1 + \frac{1}{{{p^2}}}} \right)}}{{{p^2}\left( {{p^2} + \frac{1}{{{p^2}}} - 1} \right)}}} \;dp$

$= \int {\frac{{dk}}{{{k^2} + 1}}} = {\tan ^{ - 1}}(k) + c$$ $\; \; \; \; \; \; \; \; \left( {p - \frac{1}{p} = k,\;\left( {1 + \frac{1}{{{p^2}}}} \right)\;dp = dk} \right)$$

$= {\tan ^{ - 1}}\;\left( {p - \frac{1}{p}} \right) + c = {\tan ^{ - 1}}(\tan x - \cot x) + c$$

Option 1)

log_e (tanx - cotx) + c

Option 2)

log_e (cotx - tanx) + c

Option 3)

tan^-1 (tanx - cotx) + c

Option 4)

none of these

Posted by

Plabita

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