Evaluate ∫ tan5xsec3xdx

1.	Evaluate ∫ tan5xsec3xdx
Answer» ∫ tan⁵xsec³xdx In this integral we will use the formula, 1+ tan²x = sec²x Then equation will be transform in below form- I = ∫tan⁵x sec²x secx dx = ∫secx tan⁵x sec²x dx Now, Put tan x = t which means sec²xdx = dt, I = \(\int t^5.\sqrt{1+t^2}\) dt In this above integral, Put 1+t²= k² that is mean tdt = kdk I = ∫(k⁴ + 1- 2k) k² dk = ∫ (k⁶ + k² – 2k³)dk = \(\frac{k^7}{7}\)+ \(\frac{k^3}{3}\)- \(\frac{k^4}{2}\) Now, Put the value of k = (1+t²) = sec²x in above equation- I = \(\frac{sec^{14}x}{7}\)+ \(\frac{sec^{6}x}{3}\)-\(\frac{sec^{8}x}{2}\)

Answer»

∫ tan⁵xsec³xdx

In this integral we will use the formula,

1+ tan²x = sec²x

Then equation will be transform in below form-

I = ∫tan⁵x sec²x secx dx

= ∫secx tan⁵x sec²x dx

Now,

Put tan x = t which means sec²xdx = dt,

I = \(\int t^5.\sqrt{1+t^2}\) dt

In this above integral,

Put 1+t²= k² that is mean tdt = kdk

I = ∫(k⁴ + 1- 2k) k² dk

= ∫ (k⁶ + k² – 2k³)dk

= \(\frac{k^7}{7}\)+ \(\frac{k^3}{3}\)- \(\frac{k^4}{2}\)

Now,

Put the value of k = (1+t²) = sec²x in above equation-

I = \(\frac{sec^{14}x}{7}\)+ \(\frac{sec^{6}x}{3}\)-\(\frac{sec^{8}x}{2}\)

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