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ANSWER :—

Given Expression,

\displaystyle \sf \: l = \int \big( 3 <klux>SIN</klux>(a) - 4 cos(a) \big)da

The above expression can be REWRITTEN as :

\displaystyle \sf \: l = \int3 sin(a). da - \int4cos(a).da

Consider the above expression to be of the FORM :

\sf \: l = l_1 - l_2

Firstly,

\displaystyle \: \sf \: l_1 = \int3 \: sin(a)da \\ \\ \displaystyle \longrightarrow \: \sf l_1 = \:3 \int \: sin(a).da \\ \\ \longrightarrow \boxed{ \boxed{ \sf \: l_1 = - 3cos(a) + c}}

Also,

\displaystyle \sf \: l_2 = \int 4 cos(a).da \\ \\ \longrightarrow \displaystyle \sf l_2= 4 \int cos(a)da \\ \\ \longrightarrow \boxed{\boxed{\sf l_2 = 4sin(a) + c}} \:

Integral of :

sin x = - cos x + c

cos x = sin x + c

Thus,the integral would be :

\sf \: \therefore \: , I = - [4sin(a) + 3cos(a)] + C


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